1. OSMOLALITY & OSMOLARITY
TONICITY, OSMOTICITY,
OSMOLALITY & OSMOLARITY
CHAPTER 18 PP

2. NAVIGATING YOUR A A B C D

3. A A
NAVIGATING YOUR A B C D

4. STRUCTURES
MOLECULAR
WEIGHTS
EQUIVALENCY WEIGHT
EQUATION
ELECTROLYTES OR NONE

5. A PENCILE COST $ 0.10
ERASER;......MILLIONS

7. FUNCTIONS
e.g.
TEMPERATURE;
HEART RATE;
BLOOD PRESSURE.

8. COLLIGATIVE PROPERTIES
OSMOTIC PRESSURE
BOILING POINT
VAPOR PRESSURE
FREEZING POINT

9. C A B

10. Isotonic Solutions
contain the same concentration of solute as an another solution (e.g. the cell's cytoplasm). When a cell is placed in an isotonic solution, the water diffuses into and out of the cell at the same rate. The fluid that surrounds the body cells is isotonic.

23. The Osmosis definition :
Osmosis is the passage of water from a region of high water concentration through a semi-permeable membrane to a region of low water concentration.

28. the number of particles.
Osmotic pressure
is related to
the number of particles.

29. The pressure required to prevent the movement of molecules of H O
is the osmotic pressure. 2

30. ? FACT THE OSMOTIC PRESSURE OF OUR BODY FLUIDS, EQUIVALENT TO
NaCl SOLUTION.
0.9%

31. WHY NaCl?
WHY ONLY 0.9%?
Where this (0.9%) came from?

32. Na &Cl are the most plentiful electrolyte in the body.
WHY NaCl?
Na &Cl are the most plentiful electrolyte in the body.
NORMAL HEALTHY HUMAN: Na mMol.L-1
Cl mMol.L K mMol.L-1 Ca mMol.L-1

33. WHY ONLY 0.9%?
FREEZING POINT
-0.52oC

34. -0.52oC WHY ONLY 0.9%? What are the relations between 0.9% and -0.52C?
FREEZING POINT
What are the relations
between 0.9% and -0.52C?

35. For NaCl Mole in gL-1.= - 1.86oC*i XgL-1. - 0.52oC

36. Wait a
where did this
equation came from
& What it means?

37. “Volumiegualidi gas nellestessecondizioniditemperatura e dipressionecontengono la stessonumerodimilecole”
Amedeo Avogadro
Count Lorenzo Romano Amedeo Carlo Avogadro diQuaregna e Cerreto, (TurinAugust 9, July 9, 1856)

38. AMEDEO AVOGADRO ( )
“Equal volumes of gases at the same temperature and pressure contain the same number of molecules”
Amedeo Avogadro

39. AVOGADRO’S NUMBER
6.02x1023

40. THE NUMBER OF PARTICLES IN A SOLUTION OF ONE Kg OF WATER.
AVOGADRO’S NUMBER
THE NUMBER OF PARTICLES IN A SOLUTION OF ONE Kg OF WATER.

41. Any nonelectrolyte
Mole will freeze at
-1.86 C

42. How about electrolyte?
FREEZING POINT
IS RELATED TO
THE NUMBER
OF PARTICLES.

43. THE WEIGHT IN g . OF A SOLUTE. EQUIVALENT TO A MOLE.
Osmol
THE WEIGHT IN g . OF A SOLUTE. EQUIVALENT TO A MOLE.

44. THE WEIGHT IN mg OF A SOLUTE. EQUIVALENT TO A mMOLE.
mOsm
THE WEIGHT IN mg OF A SOLUTE. EQUIVALENT TO A mMOLE.

45. IT IS THE AMOUNT OF A SOLUTE, WHICH WILL PROVIDE ONE AVOGADRO’S NUMBER
Osmol
IT IS THE AMOUNT OF A SOLUTE, WHICH WILL PROVIDE ONE AVOGADRO’S NUMBER

46. i concept

47. #

48. BELIEVE IT OR NOT!

49. Nonelectrolyte(1)
Boric acid
Dextrose
Glycerin
Mannitol
monks or nuns

50. For Boric acid Mole in gL-1.= - 1.86oC*i XgL-1. - 0.52oC

51. 1 Mole of anhydrous
Dextrose
180 g.
EQUIVALENT TO
1 osmole 180 g.

52. 1 mMole of anhydrous
Dextrose 180 mg.
EQUIVALENT TO
1 mosmole 180 mg.

53. WHY?
NONELECTROLYTE

54. Electrolyte

55. 180/100 = 1.8 NaCl Na + Cl i = Dissociation(80%) 20% 80% + 80%
FREEZING POINT “D”EPRESSION
100
i = Dissociation(80%)
NaCl Na + Cl
20% 80% %
180/100 = 1.8

56. NUMBER of PARTICLES 1 Mole of NaCl = 58.5 gL-1 1 Mole of NaCl= 2 Osmol
29.25g of NaCl= 1 Osmol

57. NUMBERofPARTICLES
1 mOsmol of NaCl
=29.25 mg.

58. Electro-lyte
Potassium Chloride

59. 180/100 = 1.8 KCl K + Cl i = Dissociation(80%) 20% 80% + 80%
FREEZING POINT “D”EPRESSION
100
i = Dissociation(80%)
KCl K + Cl
20% 80% %
180/100 = 1.8

60. Zinc Sulfat

61. 140/100 = 1.4 ZnSO4 Zn + SO4 i = Dissociation(40%) 60% 40% + 40%
FREEZING POINT “D”EPRESSION
100
i = Dissociation(40%)
ZnSO4 Zn + SO4
60% 40% %
140/100 = 1.4

62. Electro-lyte
Calcium Chloride

63. 260/100 = 2.6 CaCl2 Ca +2 Cl i = Dissociation 20% 80% + (2*80%)
FREEZING POINT “D”EPRESSION
i = Dissociation
CaCl2 Ca +2 Cl
20% 80% + (2*80%)
260/100 = 2.6

64. Potassium Sulfate

65. 260/100 = 2.6 K2SO4 2 K+SO4 20% (2*80%) + 80% i = Dissociation
FREEZING POINT “D”EPRESSION
i = Dissociation
K2SO4 2 K+SO4
20% (2*80%) %
260/100 = 2.6

66. Ferric Chloride

67. 340/100 = 3.4 FeCl3 Fe+3CL i = Dissociation 20% 80% + (3*80%)
FREEZING POINT “D”EPRESSION
i = Dissociation
FeCl3 Fe+3CL
20% 80% + (3*80%)
340/100 = 3.4

68. ISO-OSMOTIC or ISOTONIC
What are the differences between
ISO-OSMOTIC ---ISO-TONIC?
ISO-OSMOTIC or ISOTONIC or

69. Isotonic
Iso = the same
tonic =tone ,shap
infers to
physiological
compatibility.

70. Why?

71. ISOTONIC
HYPERTONIC
HYPOTONIC
crenation

72. What are the differences
between
58.5 g NaCl and g KCl?

73. iso-osmotic
but not
isotonic

75. BOTH BLOOD & LACRIMAL FLUID,
gL-1BORIC ACID
IS ISO-OSMOTIC WITH
BOTH BLOOD & LACRIMAL FLUID,
BUT ONLY ISOTONIC
WITH LACRIMAL FLUID.

76. Hemolysis TO Red Blood Cells
Boric Acid Causes
Hemolysis TO Red Blood Cells
WHY?
Boric Acid Pass Freely Through the RBC MembraneRegardless of Concentration.

78. Osmotic Diuretic

79. Na BIPHOSPHATE [FLEET]
LAXATIVE
MAGNESIUM SULFATE
MAGNESIUM CITRATE
GLYCERIN [RECTAL]
Na PHOSPHATE &
Na BIPHOSPHATE [FLEET]

80. LACTULOSE & CEPHULAC NON-ELECTROLYTE NONABSORBABLE DISACCHARIDE,
Class
Generic Name
Brand Name
Ammonia Detoxicants
Lactulose
CEPHULAC
NON-ELECTROLYTE
NONABSORBABLE DISACCHARIDE,
+ COLON BACTERIA LACTIC ACID
OSMOTIC PRESSURE
ACIDIFICATION SERVE AS A TRAP FOR AMMONIA BLOOD LEVEL
Rx: SYSTEMIC ENCEPHALOPATHY

81. DRY MUCOUS MEMBRANE:299-313 mOsmol kg-1
S&S of OSMOTICITY
Normal 285 mOsmol kg mOsmol kg-1
THIRSTY mOsmol kg-1
DRY MUCOUS MEMBRANE: mOsmol kg-1
WEAKNESS, DOUGHY SKIN: mOsmol kg-1

82. >330 mOsmol kg-1 DISORIENTATION POSTURAL HYPOTENSION
SEVERE WEAKNESS
FAINTING
COMA

83. if
But what 1%
O.P.

84. DisorientationCRAMPS250-233 mOsmol kg-1
HEADACH : mOsmol kg-1
DROWSINESS: mOsmol kg-1
DisorientationCRAMPS mOsmol kg-1
SEIZURES & COMA
<230 mOsmol kg-1

85. 2 Na + 10 SERUM OSMOLALITY 1.86 Na + BLOOD SUGAR + BUN +5 18 2.8
A QUIKY
2 Na + BLOOD SUGAR + BUN 20 3
THE QUIKIST
2 Na + 10

86. WHOLE MILK
TOMATO JUICE
ORANGE JUICE
ICE CREAM 1150

87. PITUITARY ANTIDIURETIC HORMONE (ADH)
SERUM OSMOCITY
PITUITARY ANTIDIURETIC HORMONE (ADH)

88. OSMOLALITYDETERMINATION

89. NaCl “E”QUIVALENT OS
“V” VALUES
 in RBC
FREEZING POINT

90. FREEZING POINT OS

91. FREEZING POINT “D”EPRESSION
NaCl 0.9%
-0.52oC

92. WHAT IS THE USE
OF ALL THESE B.S.?

93. Samaan Naphazoline HCl 0.02% [247 ,2] Purified water qs 30 mL
Zinc sulfate 0.25% Rx
Mft Isotonic solution
Samaan

94. FREEZING POINT D

95. Mole in gL-1= - 1.86oC*i XgL-1 - oC
FREEZING POINT “D”EPRESSION
Naphazoline HCl M.Wt Ions 2 k
Mole in gL-1= oC*i
XgL oC Rx ? Rx
Naphazoline HCl 0.02% [247 ,2]
Zinc sulfate 0.25%
water qs 30 mL
Mft Isotonic solution
0.02 g mL
0.2 g mL Rx

96. - 0.0027oC Naphazoline HCl 0.2gL-1 - XoC 247gL-1=-1.86oC*1.8
FREEZING POINT “D”EPRESSION
Naphazoline HCl
247gL-1=-1.86oC*1.8
0.2gL XoC oC

97. 0.0226 oC ZnSO4 2.5gL-1 - XoC 288gL-1= -1.86oC*1.4
FREEZING POINT “D”EPRESSION
ZnSO4
288gL-1= oC*1.4
2.5gL XoC oC

98. Total depression oC
0.0226oC oC

99. Subtract actual solution
freezing depression contribution from
the reference isotonic solution.
= ( ) = oC or
= = oC

100. 270 mg 0.52oC Xmg 0.4947oC NaCl needed to fill the 0.4947 C gap. o
mg NaCl

101. of contribution of @ drug
Step 1
Equation
How much depression it can a drug cause? 2 
of contribution drug 3 
0.52 C.
actual solution
freezing depression

102. NaCl “E”QUIVALENT OS
 in RBC
“V” VALUES
FREEZING POINT

103. NaCl “E”QUIVALENT
The weight of NaCl which will produce the same osmotic pressure effect as 1 g. of the drug.

104. Samaan Naphazoline HCl 0.02% Zinc sulfate 0.25%
Purified water qs 30 mL Rx
Mft Isotonic solution
Samaan

105. Mole of NaCl gL-1 * iDrug Mole of drug gL-1 * iNaCl NaCl “E”QUIVALENTk
58.5
Mole of NaCl gL-1 * iDrug
Mole of drug gL-1 * iNaCl
1.8
Naphazoline HCl M.Wt Ions 2

106. has the osmotic Pressur
“E” forNaphazoline HCl
58.5gL-1 *1.8
247gL-1 *1.8
= g NaCl
1 g of Naphazoline HCl
has the osmotic Pressur
as g of NaCl

107. Rx Naphazoline HCl 0.02 g* 30 mL/100 mL= 0.006 g Naphaz.
Zinc sulfate 0.25%
Purified water qs 30 mL Rx
Mft Isotonic solution
0.02 g* 30 mL/100 mL= g Naphaz.
g NaCl*0.006 g Naph= g NaCl
1 g Naphz

108. has the osmotic Pressure
“E” forZnSO4
58.5gL-1*1.4 =
288gL-1*1.8
1 g of ZnSO4
has the osmotic Pressure
as g of NaCl

109. RxZnSO4 0.25*30/100=0.075 g Zn SO4. 0.1579*0.075= 0.0118 g NaCl
Naphazoline HCl 0.02%
Zinc sulfate 0.25%
Purified water qs 30 mL Rx
Mft Isotonic solution
0.25*30/100=0.075 g Zn SO4.
0.1579*0.075= g NaCl

110.  0.006 g Naphaz.HCl = 0.0014 g NaCl 0.075 g Zn SO4 = 0.0118 g NaCl

111. 9 mgmL-1 NaCl*30 mL=270 mg NaCl.
If there no medication, this prescription will be isotonic with 270 mg of NaCl.

112. NaCl mg
Purified water qs 30 mL
Mft Isotonic solution Rx
Samaan

113. NaCl 270 mg
NaCl mg
NaCl mg

114. 256.8 mg NaCl with “E”
257 mg NaCl with “D”

115. Samaan Naphazoline HCl 6 mg Zinc sulfate 75 mg Sodium chloride 257 mg
Purified water qs 30 mL Rx
Samaan

116. NaCl “E”QUIVALENT
“V” VALUES OS
 in RBC
FREEZING POINT

117. V
olumeof water to be added to a specified weight of drug to prepare an isotonic solution.

118. Samaan Naphazoline HCl 0.02% Zinc sulfate 0.25%
Purified water qs 30 mL Rx
Mft Isotonic solution
Samaan

119. value is based on “E” value

120. Naphazoline HCl 0.02% Rx
Samaan

121. has the osmotic Pressure
“E” for Naphazoline HCl
58.5gL-1 *1.8
247gL-1 *1.8
= g NaCl
1 g of Naphazoline HCl
has the osmotic Pressure
as g of NaCl
= g NaCl* 0.006g naph.HCl= g NaCl

122. Dissolve 6 mg of Naph.HCl in 0.16 of water, the sol. will be isotonic.
0.006 g N.HCl= g NaCl
g NaCl* 1 ml= mL H2O
0.009 gNaCL
Dissolve 6 mg of Naph.HCl in 0.16 of water, the sol. will be isotonic. mL

123. Zinc sulfate 0.25% Rx
Samaan

124. “E” for ZnSO4 58.5gL-1*1.4 288gL-1*1.8 = 0.1579 g NaCl
1 g of ZnSO4 has the osmotic Pressure
as g of NaCl
= g NaCl* g ZnSO4
= g NaCl

125. g NaCl/0.009 g NaCl=
1.31 mL
If you dissolve 75 mg of ZnSO4 in 1.31 mL of H2O, the sol. will be isotonic.

126. mL mL
1.4704mL

127. Dissolve 6 mg of naph. HCl& 75 mg ZnSO4 in 1
Dissolve 6 mg of naph.HCl& 75 mg ZnSO4 in 1.47 mL water, and qs to 30 mL with isotonic 0.9% NaCl.

128. PROBLEM! WHAT PROBLEM?

129. 1 1
Chapter

130. P1
BORIC ACID “E” 0.52
Isotonic NaCl solution contains 0.9% w/v NaCl . If the “E” value of boric acid is 0.52, calculate the % strength (w/v) of an isotonic solution of boric acid.

131. 1 g. of boric acid has the same osmotic pressure as 0.52 g. of NaCl.
BORIC ACID “E” 0.52
1 g. of boric acid has the same osmotic pressure as 0.52 g. of NaCl.

132. 1 g B.A. * 0.9 g 0.52 g IF 0.9% NaCl IS ISOTONIC X g BORIC ACID =
1 g OF BORIC ACID = 0.52 g NaCl
X g OF BORIC ACID= 0.9 g NaCl
X g BORIC ACID =
1 g B.A. * 0.9 g
= 1.73 g%
0.52 g

133. Mole gL-1= - 1.86oC*i x gL-1 - 0.52oC 61.8gL-1= - 1.86oC*1.0

134. 17.27 g mL
x g mL
1.727g%

135. IF NaCl DISSOCIATING AT 90%. CALCULATE:- A- DISSOCIATION FACTOR;P2
IF NaCl DISSOCIATING AT 90%. CALCULATE:-
A- DISSOCIATION FACTOR;
B- FREEZING POINT OF MOLAL SOLUTION.

136. A-
i =Dissociation(2)
NaCl Na + Cl
10% 90% %
190/100 = 1.9

137. B-
- XoC= oC*1.9
X = oC

138. P3
WHAT IS THE F.P. OF 25 g IN 500 mL DEXTROSE?
D5W

139. = -0.5167oC Mole gL-1= - 1.86oC*i [ gL-1] - XoC 180 gL-1= - 1.86oC*1

140. P4 PROCAINE HCl [M.Wt 273; 2-ION] DISSOCIATING AT 80%.
A- DISSOCIATION FACTOR,
B-”E”
C- F.P. FOR A MOLAL SOLN.

141. 180/100 =1.8 20% 80% + 80% A- DISSOCIATION FACTOR,
PROCAINE HCl PROCAINE + HCl
20% 80% %
180/100 =1.8

142. B- “E”
58.5gL-1*1.8 =
273gL-1*1.8
1 g Procaine HCl= g NaCl

143. C- F.P. FOR A MOLAL SOLUTION
= oC*i
= oC *1.8
= oC

144. P5 The freezing point of a molal solution of a nonelectrolyte is
-1.86°C. What is the freezing point of a 0.1 % solution of
zinc chloride (M.Wt. 136), dissociating 80%?

145. F.P. OF 0.1%ZnCl2 [MWt=136,3]
Mole gL-1= oC*i
F.P.D=
[gL-1] XoC
136 gL-1= oC*2.6
1gL xoC
= oC

146. P6
F.P. of 5% boric acid is -1.55oc.
HOW many g. of boric acid should be used to prepare one L of an isotonic sol.?

147. Mole gL-1= oC*i
XgL oC
50 gL = oC*1
XgL oC
gL-1

148. Mole gL-1= oC*i
XgL oC
6I.8 gL-1= oC*1
XgL oC
gL-1

149. P7 Eph.SO4 429,3 How many mg. of NaCl? Rx
Ephedrine sulfate mg Sodium Chloride q.s.
Purified water ad mL
Make isotn.sol.
Sig. Use as directed
[429,3]
How many mg. of NaCl?

150. Rxof contribution of @ drug
Step  2  1
R How much NaCl can make the whole Rx isotonic?
Rxof contribution drug  3
actual amount of the adjustor
to make the whole Rx isotonic
R-Rx

151.  2  1 NaCl ‘E’ 3  Step = 270 mg of NaCl = 0.1969 g Nacl
9 mg NaCl* 30 mL the total volume of the Rx
1mL
= 270 mg of NaCl R 2 
58.5gL-1*2.6
= g Nacl
429gL-1*1.8 Rx
1 g of Ephedrine sulfate has the same osmotic pressure as g Nacl
0.3 g * = g NaCl= 59 mg NaCl 3 
270 mg -59 = mg NaCl

152. P8 How many g. of NaCl? RxDipivefrin HCl 0.5% [388,2]
Scopolamine HBr0.33[438,2]
SodiumChlorideq.s.
Purified water ad mL Make isotn.sol.
Sig. Use in the eyes.
How many g. of NaCl?

153. Dipivefrin HCl E
58.5gL-1*1.8
= 0.15 gNaCl
388gL-1*1.8
0.5 g mL x g mL
0.15 g * 0.15 = g
22.5 mgNaCl

154. E 0.33 g 100 mL x g. 30 mL Scopolamine HBr 58.5gL-1*1.8 438gL-1*1.8
= gNaCl
438gL-1*1.8
0.33 g mL x g mL
0.1 g * = gNaCl

155. Both drugs ‘E’ g g g

156. NaCl reference -‘E’
0.270 g
0.035 g
0.234 g

157. P9 How many g. of boric acid?
Rx Zinc sulfate Boric acidq.s. Purified water ad mL Make isotn. sol.
Sig. drop in eyes
How many g. of boric acid?

158. = 0.1579 “E” for Zinc sulfate 58.5gL-1*1.4 288gL-1*1.8
0.06* = g NaCl

159. 0.270 g 0.0095 g 0.2605 g NaCl reference - ‘E’
but Rx calls for boric acid!

160. 0.500 g Boric acid 1 g Boric acid 0.52 g NaCl
x g Boric acid g NaCl
0.500 g Boric acid

161. Boric acid isotonic reference
17.3 mg*30 mL
1 mL
=519 mg ofBoric acid to make 30mL isotonic [reference]
“EB” for Zinc sulfate
61.8gL-1*1.4
= 0.3 g B.A.
288gL-1*1.0
0.06* 0.3= g Boric acid

162. 0.500 g Boric acid 0.519 g of B.A.[reference]
0.018 g of boric acid equivalent
0.500 g Boric acid

163. P 10 Rx Cromolyn Na4% [512,2] Benzalkonium Cl. (1:10,000) [360,2]
Buffer sol q.s. Purified water ad mL
Make isotn. sol.
Sig. One drop in each eye
How many mL of the buffer solution (E = 0.30) should be used to render the solution isotonic?

164. = 0.1142 g NaCl “E” for Cromolyn Na 58.5gL-1*1.8 512gL-1*1.8
R= 9 mg NaCl* 10 mL/1 mL = 90 mg NaCl
“E” for Cromolyn Na
58.5gL-1*1.8
= g NaCl
512gL-1*1.8
0.4 g C.Na* = 0.045g NaCl

165. = 0.1625g NaCl “E” for Benzalkonium Cl. (1:10,000)[360,2] 58.5gL-1*1.8
0.001g Bz.* = g NaCl
0.09 g NaCl g NaCl
g NaCl

166. (E = 0.30) 1 g of buffer material = 0.3 g of NaCl 0.3 g of NaCl
g NaCl * 1 g of buffer material
0.3 g of NaCl
= g of buffer material
3 g of buffer material = 0.9 g of NaCl
3 % of buffer sol. = 0.9 % of NS

167. -3.0g Buffer solution 100 mL
0.147 g of buffer X mL
X mL of isotonic buffer solution =0.147 g * 100 ml/3.0 g buffer= 4.9 mL
4.9 mL of isotonic buffer solution

168. P11 Rx Dextrose,anhydrous 2.5% NaCl q.s.
Sterile water for injection ad 1000mL Label: isotonic Dextrose & Saline Solution.
How many g of NaCl needed?

169. 25 g.* 0.18= 4.51g NaCl = 0.18 g NaCl 4.5 g NaCl 58.5gL-1*1
“E” for anhydrous Dextrose
R = 9.0 g NaCl*1L/1L= 9.0 g NaCl
58.5gL-1*1
= 0.18 g NaCl
180gL-1*1.8
25 g.* 0.18= 4.51g NaCl
9.0 g NaCl
-4.5 g NaCl.
4.5 g NaCl

170. P Rxsol.Silver Nitrate 0.5% 15.0 Why not to use NaCl as adjustor? 12
Make isoton. sol.
Sig. For the eyes.
How many g of KNO3 needed?
Why not to use NaCl as adjustor?

171. Reference of KNO3
Mole gL-1= oC*i
XgL oC
101gL-1= oC*1.8
XgL oC
15.69gL-1

172. 0.235 g of KNO3 to fill up this prescription without any medication.
Reference of KNO3
15.69g mL
x g mL
0.235 g of KNO3 to fill up this prescription without any medication.

173. “EKNO3” for Silver Nitrate
101gL-1 *1.8
=0.5941gKNO3
170gL-1 *1.8
0.075 g AgNO3* = gKNO3
0.235 g of KNO3 Reference
g of KNO3 Rx
0.190g of KNO3

174. P13 Rx Cocaine HCl 0.15[340,2] NaCl q.s. Purified Water ad 15
Make isoton. sol.
Sig. One drop for the left eye.
How many g of NaCl needed?

175. 58.5gL-1*1.8 = 0.172g NaCl 340gL-1*1.8 “E” for Cocaine HCl
R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl
“E” for Cocaine HCl
58.5gL-1*1.8
= 0.172g NaCl
340gL-1*1.8
0.15 g C.HCl* 0.172= g NaCl
R-Rx= g g= 0.109gNaCl

176. P14
Rx Cocaine HCl [340,2]
Eucatropine HCl 0.6 [328,2] Chlorobutanol [177,1]
NaCl qs
Purified Water ad 30
Make isoton. sol.
Sig. For the eyes.

177. = 0.172g NaCl “E” for Cocaine HCl 0.6 g CHCl.* 0.172= 0.1032 g NaCl
R=0.009g NaCl * 30 mL/1 mL =0.27 g NaCl
“E” for Cocaine HCl
58.5gL-1*1.8
= 0.172g NaCl
340gL-1*1.8
0.6 g CHCl.* 0.172= g NaCl

178. = 0.178 58.5gL-1*1.8 328gL-1*1.8 “E” for Eucatropine HCl
0.6 g Euc.* 0.178= g NaCl

179. = 0.1836 58.5gL-1*1.0 177gL-1*1.8 “E” for Chlorobutanol
0.1 g Chl.* = g NaCl

180. Cocaine HCl0.1032 g Eucatropine HCl 0.1070 g Chlorobutanol 0.0184 g
0.270 g g = g

181. P15 RxTetracaine HCl 0.1 [301,2] Zinc sulfate 0.05 [288,2]
Boric acidqs
Purified Water ad 30
Make isoton. sol.
Sig. For the eyes.
How many g. Boric acid needed?

182. = 0.3695 g Boric Reference “Eb” for Tetracaine HCl
17.3 g*30 mL= g Boric
1000 mL
“Eb” for Tetracaine HCl
61.8gL-1*1.8
= g Boric
301gL-1*1.0
0.1 g Tetr.* = g Boric

183. “Eb” for Zinc sulfate 61.8gL-1*1.4 288gL-1*1.0 = 0.3004 g Boric
0.05 g Zn.* = g Boric

184. 0.467 g Boric acid 0.0369 g 0.0150 g 0.0519 g Tetracaine HCl
Zinc sulfate g g
0.519 g Boric acid Reference
0.0519g Boric acid Equivalent
0.467 g Boric acid

185. P16 Rx Sol. HomatropineHBr 1% 15 [356,2] Boric acid q.s.
Make isoton. sol.
Sig. For the eyes.
How many g. Boric acid needed?

186. = 0.3125 17.3 g*15 mL/1000 mL Boric acid Reference 61.8gL-1*1.8
0.2595g Boric acid
“Eb” for HomatropineHBr
61.8gL-1*1.8 =
356gL-1*1.0
x g = 0.15 g

187. 0.212 g Boric Acid Boric acid Equivalent 0.2595 g Boric acid reference
0.15 g Homa.* = g Boric
g Boric acid reference
g Boric acid Equivalent
0.212 g Boric Acid

188. P17 Rx Procaine HCl 0.1% [273,2] NaCl q.s. How many g. NaCl needed?
SterileWater for Inj. ad
Make isoton. sol.
Sig. For Injection.
How many g. NaCl needed?

189. 0.6857 g NaCl “E” for Procaine HCl 58.5gL-1*1.8 = 0.2142 g NaCl
R=0.009g NaCl * 100 mL/1 mL =0.9 g NaCl
“E” for
Procaine HCl
58.5gL-1*1.8
= g NaCl
273gL-1*1.8
1 g Proc.* = g NaCl
0.9 g NaCl reference
g NaCl Equival
g NaCl

190. P18 Rx Phenylephrine HCl 1%[204,2] Chlorobutanol 0.5%[177,1] NaCl q.s.
Purified Water ad
Make isoton. sol.
Sig. Use as directed
How many mL NSS needed?

191. = 0.2867 g NaCl 58.5gL-1*1.8 204gL-1*1.8 “E” for Phenylephrine HCl
R=0.009g NaCl * 15 mL/1 mL =0.135 g NaCl
“E” for Phenylephrine HCl
58.5gL-1*1.8
= g NaCl
204gL-1*1.8
0.15 g Phen.* = g NaCl

192. “E” for Chlorobutanol = 0.1836 g NaCl 177gL-1*1.8 58.5gL-1*1.0
0.075 g Ch.* =
Phenylephrine HCl + Chlorobutanol
0.043 g NaCl g NaCl = g NaCl
R-Rx= =0.078 g NaCL
0.078 g NaCl*100 mL
0.9 g NaCl
8.69 mL N.S.

193. P19 Rx Oxymetazoline HCl 0.5% [297,2] Boric acid sol. qs
Purified Water ad
Make isoton. sol.
Sig. For the nose, as decongestant
How many mL of 5% boric acid solution needed?

194. R = 0.3745 g B.A “Eb” for 0.075 g Oxy* 0.3745= 0.028 g B.A.
17.3 g*15 mL = g B.A mL R
“Eb” for
Oxymetazoline HCl
61.8gL-1*1.8
= g B.A
297gL-1*1.0
0.075 g Oxy* = g B.A.

195. 4.63 mL of 5% Boric acid solution.
g B.acid reference
0.028 g B.acidEquival
g Boric acid
g B.A.*100 ml/5 g B.A.
4.63 mL of 5% Boric acid solution.

196. P20 How many g of Dextrose needed? Rx Ephedrine HCl 0.55 [202,2]
Chlorobutanol [177,1]
Dextrose qs
Rose Water ad
Make isoton. sol.
Sig. Nose drop.
How many g of Dextrose needed?

197. 2.516 g Dex. Ref. 180 g L-1 = 1.86 *1 x g L-1 -.52oC 50.32 gL-1
50.32 g*0.05 L L
2.516 g Dex. Ref.

198. = 1.6 180gL-1*1.8 202gL-1*1.0 “Ed” for Ephedrine HCl
0.5 g Eph* 1.6= 0.80g Dex.

199. = 1.0169 180gL-1*1.0 177gL-1*1.0 “Ed” for Chlorobutanol
0.25 g Ch* = g Dex.

200. 1.056 g Dex. 2.516 g Dex. Ref. 1.056 g Dex. Equival 1.4598 g Dex.
Ephedrine HCl 0.80g Dex.
Chlorobutanol g Dex.
1.056 g Dex.
2.516 g Dex. Ref.
1.056 g Dex. Equival
g Dex.

201. P21 Naphzoline HCl 1% [247,2] Sodium Chloride qs
Purified Water ad mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many g of NaCl needed? Using Freezing point method.

202. 247 gL-1= oC*1.8
10gL xoC oC
0.52-
( oC)= oC

203. 0.1995 g NaCl 0.270 g NaCl Reference 0.27 g freezes at -0.52oC
x g NaCl oC
g NaCl

204. P22 Rx Oxytetracycline HCl 0.05 [497,2] Chlorobutanol 0.1 [177,1]
NaCl qs
Purified Water ad mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many mg of NaCl needed?

205. = 0.1177 58.5gL-1*1.8 497gL-1*1.8 “E” for Oxytetracycline HCl
0.05 g Oxy* = g NaCl

206. = 0.1836 58.5gL-1*1.0 177gL-1 *1.8 “E” for Chlorobutanol
0.1 g Chl* = g NaCl

207. Oxytetracycline HCl 0.0058g Chlorobutanol 0.0183g
0.270 g NaCl Reference g
0.2469g NaCl

208. P23 Rx
Tetracaine HCl % [301,2]
[iso]Sol. Epineph. Bitart.10.0 [333,2]
Boric acidqs
Purified Water ad mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many g of Boric acid needed?

209. 17.3 g*20 mL/1000 mL
0. 35 g B.A. for 20 mL Reference
“Eb” for Tetracaine HCl
61.8gL-1*1.8 =
301gL-1*1.0
0.15 g Tet.* = g B.A.

210. 0. 35 g B.A. for 20 mL Reference
g B.A.
0.2946g Boric acid

211. P24 Anhyd.NaH2PO4 5.6 g[120,2] Anhyd.Na2HPO4 2.84 g[142,3] NaCl qs
Purified Water ad 1000 mL Label: Isotonic buffer sol.,pH6.5
How many g of NaCl needed?

212. = 0.49 58.5 gL-1*1.8 120 gL-1*1.8 “E” for Anhyd.NaH2PO4
5.6 g Mono* 0.49= g NaCl

213. “E” for Anhyd.Na2HPO4 = 0.595 2.84 g Di* 0.595= 1.69 g NaCl
58.5 gL-1*2.6
= 0.595
142 gL-1*1.8
2.84 g Di* 0.595= 1.69 g NaCl

214. 4.566 g NaCl Anhyd.NaH2PO4 Anhyd.Na2HPO4 9.0 g NaCl Reference

215. P25
How many g of anhydrous Dextrose needed in preparing 1 L of a 0.5% isotonic Ephedrine Sulfate [429,3] Nasal spray?

216. “Ed” for Ephedrine Sulfate
180gL-1 *2.6
=1.0909
429 gL-1*1.0
5.0 g Ephd.* 1.09= 5.45 g Dex.
50 gL-1 Anhydrous Dex.Ref gL-1 Dex.
44.54 g Dextrose

217. P26 Ephedrine Sulfate 1% [429,3] Chlorobutanol 0.5%[177,1]
Purified Water ad
Make isoton. sol. & Buffer at 6.5 Sig. Nose drop.
How many mL of a buffer & mL of water should be used?

218. “E” for Ephedrine Sulfate
58.5gL-1*2.6 =
429gL-1*1.8
1.0 g Eph* = g NaCl

219. 100 mL H2O 0.9 g NaCl V-Value for Ephedrine Sulfate x mL 0.1969 g NaCl
21.87 mL of water will make 1 g of Ephedrine Sulfate isotonic.

220. “E” for Chlorobutanol = 0.1836 58.5gL-1*1.0 177gL-1*1.8
0.5 g Ch* = g NaCl

221. 100 mL H2O 0.9 g NaCl V-Value For Chlorobutanol x mL 0.0918 g NaCl
x mL= mL of water will make 0.5 g of chlorobutanol isotonic.

222. Ephedrine Sulf. 21.87 mL of water
Chlorobutanol mL of water
32.07 mL of water
67.93 mL of isotonic buffer solution.

223. P27 Oxytetracycline HCl 0.5% [497,2] [iso]Tetracaine HCl Sol. 2%15 ml
NaClqs
Purified Water ad mL
Make isoton. sol.
Sig. Use as directed in the eye.
How many mL of NSS needed?

224. 1.8 “E” for Oxytetracycline HCl 58.5gL-1* 497gL-1*1.8 = 01177g NaCl.
0.15 g Oxy* =

225. 0.135 g NaCl Reference in only 15 mL 0.0176 g NaCl Equivalent
0.9 g NaCl mLx mL
g. NaCl
13.0 mL of NSS

226. Determine if the following commercial products
Determine if the following commercial products are Hypotonic, isotonic, or Hypertonic:
a- An ophthalmic sol. 40 mgmL-1 of Cromolyn Sodium [512,2]& 0.01% of Benzalkomiun chloride [360,2]
in purified water.

227. “E” for CromolynSodium 512gL-1 = -1.86oC*1.8 40gL-1 = -XoC
HYPER
ISO-T.
HYPO
>-0.52
-0.52
<-0.52
-0.26oC

228. A parenteral infusion containing 20% (w/v) of mannitol.
FP ”D” Mannitol
182gL-1 = -1.86oC*1.0
200gL-1 = -XoC
HYPER
ISO-T.
HYPO
>-0.52
-0.52
<-0.52

229. 180gL-1 = -1.86oC*1.0 50gL-1 = -XoC ISO-T. HYPO >-0.52 -0.52
A 500-mL large volume parenteral containing D5W (5% w/v of anhydrous dextrose in sterile water for injection).
180gL-1 = -1.86oC*1.0
50gL-1 = -XoC
HYPER
ISO-T.
HYPO
>-0.52
-0.52
<-0.52
-0.519

230. A FLEET saline enema containing 19 g of monobasic sodium phosphate (monohydrate) and 7 g of dibasic sodium phosphate (heptahydrate) in 118 mL of aqueous solution.

231. Monobasic sodium phosphate (monohydrate) (138, 2)
138gL-1 = -1.86oC*1.8
161gL-1 = -XoC
HYPER
ISO-T.
HYPO
>-0.52
-0.52
<-0.52
-3.9

232. Dibasic sodium phosphate (heptahydrate) (268,3)
268gL-1 = -1.86oC*2.6
59.32gL-1 = -XoC
HYPER
ISO-T.
HYPO
>-0.52
-0.52
<-0.52
-1.07

233. For agents having the following sodium chloride equivalents, calculate the percentage concentration of an isotonic solution:
(a) 0.20
0.90%
0.20
= 4.5%
0.90%
0.32
(b) 0.32
= 2.81%
(c) 0.61
0.90%
0.61
= 1.48%

234. P30 1 g * 3 mL/100 mL = 0.3 g fentanyl citrate
How many mL each of purified water and an isotonic sodium chloride solution should be used to prepare 30 mL of a 1% w/v isotonic solution of fentanyl citrate
(E = 0.11)?
1 g * 3 mL/100 mL = 0.3 g fentanyl citrate
0.3 g fentanyl citrate* 0.11= g NaCl
0.033 g NaCl* 100 mL/0.9 g NaCl= 3.66 mL H2O
30 mL mL H2O= mL N.S.

235. Calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:
(a) Antipyrine [ 188, 1]
58.5*1/(188*1.8) = g * 0.3 = g NaCl
0.051 g NaCl* 100 mL / 0.9 g NaCl = mL H2O

236. Using the E values in Table 11
Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:
(b) Chlorobutanol [177, 1]
58.5*1/(177*1.8) = g * 0.3 = g NaCl
g NaCl* 100 mL / 0.9 g NaCl = mL H2O

237. (c) ephedrine sulfate [429, 3]
Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:
(c) ephedrine sulfate [429, 3]
58.5*2.6/(429*1.8) = g * 0.3 = g NaCl
0.059 g NaCl* 100 mL / 0.9 g NaCl = mL H2O

238. Using the E values in Table 11
Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:
(d) silver nitrate [170, 2]
58.5*1.8/(170*1.8) = g * 0.3 = g NaCl
0.103 g NaCl* 100 mL / 0.9 g NaCl = mL H2O

239. Using the E values in Table 11
Using the E values in Table 11.1, calculate the number of mL of water required to make an isotonic solution from 0.3 g of each of the following:
(e) zinc sulfate [288, 2]
58.5*1.4/(288*1.8) = g * 0.3 = g NaCl
g NaCl* 100 mL / 0.9 g NaCl = mL H2O