1. 02 – CONCENTRATION - MOLARITY CHEMISTRY 30 – UNIT 2 – SOLUBILITY – CH. 16 IN TEXT

2. CONCENTRATION THE WORD “CONCENTRATION” REFERS TO HOW MUCH SOLUTE IS DISSOLVED IN THE SOLUTION. YOU GET THE IDEA: DISSOLVING TWO CUPS OF SUGAR IN ONE LITRE OF WATER WILL PRODUCE A MORE CONCENTRATED SOLUTION THAN DISSOLVING ONE CUP OF SUGAR IN ONE LITRE OF WATER WOULD.

3. CONCENTRATION “CONCENTRATION” CAN BE MEASURED IN A VARIETY OF WAYS. BY FAR THE MOST USEFUL IS MOLARITY. THE MOLARITY OF A SOLUTION IS CALCULATED BY TAKING THE MOLES OF SOLUTE AND DIVIDING BY THE LITRES OF SOLUTION.

4. MOLARITY

5. EXAMPLE #1 SUPPOSE WE HAD 1.00 MOLE OF SUCROSE (IT'S ABOUT 342.3 GRAMS) AND PROCEEDED TO MIX IT INTO SOME WATER. IT WOULD DISSOLVE AND MAKE SUGAR WATER. WE KEEP ADDING WATER, DISSOLVING AND STIRRING UNTIL ALL THE SOLID WAS GONE. WE THEN MADE SURE THAT WHEN EVERYTHING WAS WELL-MIXED, THERE WAS EXACTLY 1.00 LITER OF SOLUTION. WHAT WOULD BE THE MOLARITY OF THIS SOLUTION?

6. EXAMPLE #1 MOLARITY = 1.00 MOL 1.00 L THE ANSWER IS 1.00 MOL/L. NOTICE THAT BOTH THE UNITS OF MOL AND L REMAIN. NEITHER CANCELS. A REPLACEMENT FOR MOL/L IS OFTEN USED. IT IS A CAPITAL M. SO IF YOU WRITE 1.00 M FOR THE ANSWER, THEN THAT IS CORRECT.

7. EXAMPLE # 2 SUPPOSE YOU HAD 2.00 MOLES OF SOLUTE DISSOLVED INTO 1.00 L OF SOLUTION. WHAT'S THE MOLARITY? MOLARITY = 2.00 MOL 1.00L THE ANSWER IS 2.00 M. NOTICE THAT NO MENTION OF A SPECIFIC SUBSTANCE IS MENTIONED AT ALL. THE MOLARITY WOULD BE THE SAME. IT DOESN'T MATTER IF IT IS SUCROSE, SODIUM CHLORIDE OR ANY OTHER SUBSTANCE. ONE MOLE OF ANYTHING CONTAINS 6.02 X 10 23 UNITS.

8. EXAMPLE # 3 WHAT IS THE MOLARITY WHEN 0.750 MOL IS DISSOLVED IN 2.50 L OF SOLUTION? MOLARITY = 0.750 MOL 2.50 L THE ANSWER IS 0.300 M.

9. EXAMPLE # 4 WE CAN CALCULATE THE MOLARITY OF A SOLUTION WHEN GIVEN THE AMOUNT OF SOLUTE IN GRAMS. WE CONVERT FROM GRAMS TO MOLES, AND THEN PROCEED AS IN THE PREVIOUS EXAMPLES. WHAT IS THE MOLARITY OF A 2.00L SOLUTION THAT HAS 102.44G NaCl DISSOLVED IN IT? STEP ONE: CONVERT GRAMS TO MOLES. STEP TWO: DIVIDE MOLES BY LITERS TO GET MOLARITY.

10. GIVEN ANY TWO OF THE VALUES FROM THE MOLARITY CALCULATION, WE CAN EASILY CALCULATE THE THIRD EXAMPLE #5 – HOW MANY GRAMS OF NaOH ARE CONTAINED IN 1.45 L OF A 2.25 M SOLUTION? STEP ONE: DETERMINE HOW MANY MOLES OF NaOH ARE PRESENT IN THE SOLUTION. STEP TWO: CONVERT FROM MOLES TO GRAMS USING THE FORMULA WEIGHT OF NaOH.

11. EXAMPLE # 6 WHAT VOLUME (IN mL) OF A 0.20 M SOLUTION OF NaOH CONTAINS 1.00 G OF NaOH? STEP ONE: DETERMINE HOW MANY MOLES ARE IN 1.00G OF NaOH. STEP TWO: USE YOUR MOLE VALUE ALONG WITH YOUR MOLARITY TO SOLVE FOR VOLUME

12. HERE’S A (SEEMINGLY) MORE COMPLICATED PROBLEM: EXAMPLE #7 – A 0.050 M SOLUTION OF GLYCERINE, C 3 H 8 O 3, AND A 0.050 M SOLUTION OF LYCINE, C 5 H 11 NO 2, ARE PREPARED. WHICH SOLUTION CONTAINS THE MOST DISSOLVED MOLECULES PER LITRE?

13. A (SEEMINGLY) COMPLICATED PROBLEM)  THIS LOOKS HARD, BUT IT’S VERY EASY. REMEMBER THAT M MEANS MOLES/LITRE. SINCE WE’RE WORKING WITH ONE LITRE VOLUMES, EACH SOLUTION WILL THEREFORE HAVE 0.050 MOLES OF SOLUTE. EACH SOLUTION WILL ALSO THEREFORE HAVE THE SAME NUMBER OF MOLECULES: MULTIPLY 0.050 MOLES BY AVOGADRO’S NUMBER.

14. STOICHIOMETRY WITH SOLUTIONS EXAMPLE #8 – HOW MANY GRAMS OF COPPER WILL REACT TO COMPLETELY REPLACE SILVER FROM 208 ML OF 0.100 M SOLUTION OF SILVER NITRATE, AgNO 3 ? YOUR PRODUCT WILL BE COPPER (II) NITRATE. USE THE SAME STOICHIOMETRY STEPS AS ALWAYS: 1. GET A BALANCED CHEMICAL EQUATION 2. TAKE YOUR INFO AND GET TO MOLES 3. USE THE MOLE RATIOS TO REBALANCE THE EQUATION 4. GO WHERE THE QUESTION ASKS