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MOLARITY Molarity expresses the concentration of a solution in number of moles of solute per liter of solution. Mathematically; M=n/V where M=molarity (in moles solute/liter solution) n=number of moles of solute V=volume of solution (in liters)

SAMPLE PROBLEM # 1 A solution of sodium hydroxide was prepared by dissolving 10.0 grams of sodium hydroxide (NaOH) in distilled water making the volume up to 500 cm 3. What is the molarity of the NaOH solution? Atomic masses: Na=23; O=16, H=1 GIVEN: mass of solute (NaOH)= 10.0 grams molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles V of NaOH solution= 500cm 3 (1L/1000cm 3 )=0.50L BACK NEXT

Solution to SP # 1 GIVEN: mass of solute (NaOH)= 10.0 grams molar mass of NaOH=1(23)+1(16)+1(1) =40g/mole n (moles of NaOH)= 10.0g(1mol/40g)=0.25moles V of NaOH solution= 500 cm 3 (1L/1000cm 3 )=0.50L SOLUTION: M(NaOH solution)=n(moles NaOH)/V(volume solution in L) =0.25moles /0.50L = 0.5 M or 0.5 moles/liter A 0.5 M solution, means there are 0.5 moles of NaOH in 1 liter of NaOH solution NEXT BACK

SAMPLE PROBLEM # 2 Calculate the number of moles copper sulfate in a 250 milliliter (mL) of 0.1 molar copper sulfate (CuSO 4 ) solution. GIVEN: M =0.1 moles/L n (moles of CuSO 4 )= ? V of CuSO 4 solution= 250mL(1L/1000mL)=0.25L NEXT BACK

Solution to SP # 2 GIVEN: M =0.1 moles/L n (moles of CuSO 4 )= ? V of CuSO 4 solution= 250mL(1L/1000mL)=0.25L SOLUTION: n(moles CuSO 4 )=(V of CuSO 4 solution in L)x (M of CuSO 4 solution) =(0.25 L) (0.1 moles/ L) = 2.5 moles CuSO 4 There are 2.5 moles of CuSO 4 in a 250 mL of 0.1 M CuSO 4 solution NEXT BACK

SAMPLE PROBLEM # 3 1.What is the volume of a 0.50 molar potassium bromide (KBr) solution containing 1.5 moles of potassium bromide (KBr)? GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in 1L KBr solution n =1.5 moles KBr V of KBr solution=? NEXT BACK

GIVEN: M =0.5 moles/L means there are 0.5 moles KBr in 1L KBr solution n =1.5 moles KBr V of KBr solution=? Solution to SP # 3 SOLUTION: V of KBr solution in L) =n(moles KBr)/( M of KBr solution) (1.5 moles) (0.5 moles/ L) = = 3.0 L KBr NEXT BACK

3. A 250 milliliter bottle of vinegar reads 5% acidity (5 g acetic acid in 100 gram vinegar).Calculate the concentration of acetic acid in moles per liter. (1mole CH 3 COOH=60 grams) ( 100mL) (60g acetic acid) SAMPLE PROBLEM # 4 GIVEN: V of solution =250 mL =0.208 moles acetic acid n acetic acid=250mL (5g acetic acid)(1mole acetic acid) M=0.208 moles 0.25 L M=0.83 moles/liter NEXT BACK

THINK PAIR SHARE 1. Find your pair. 2.Discuss how you are going to solve the solutions at work problems 3. Write your answers in “SOLUTIONS AT WORK” SHEET #1 4. Post your outputs in the “output bulletin” provided in our classroom. 5. Write your comments on each output posted. NEXT BACK

SOLUTIONS AT WORK Sheet #1 1.A solution of glucose (C 6 H 12 O 6 ) contains 300 grams in 1.0 liter of solution. What is the concentration of the glucose solution in moles per liter? Atomic masses: C=12, H=1, O=16 NEXT BACK

SOLUTIONS AT WORK Sheet #1 2. A 3.0 grams of magnesium hydroxide, Mg(OH) 2 is dissolved in water to produce a 0.2 molar magnesium hydroxide solution. What is the volume of the solution prepared? Atomic masses: Mg=24, H=1, O=16 NEXT BACK

There are times when glucose solution must be introduced into the bodies of patient. Why is it dangerous for this solution to be highly concentrated? Assignment: Answer the following briefly: NEXT HOME BACK