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Dilution/Colligative property review. 1. A 3.5 M solution of HCl has a volume of 200.0 mL. Calculate the new molarity when it is diluted with 300.0 mL of water. M 1 V 1 = M 2 V 2 (3.5 M) ( 200 ml) = X ( 500 mL) X = 1.4 M Don’t forget that V 2 is 200 + 300.
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1. A 3.5 M solution of HCl has a volume of 200.0 mL. Calculate the new molarity when it is diluted with 300.0 mL of water.
M1V1 = M2V2(3.5 M) ( 200 ml) = X ( 500 mL)X = 1.4 MDon’t forget that V2 is 200 + 300
An acid solution is diluted with 250 mL of water so that the new concentration of the solution is 2.25 M. If the initial concentration of the acid was 1.15 M, what was its original volume.
(2.25 M) (X) = (1.15 M)(X + 250ml)2.25 X = 1.15 X + 287.5X = 261.4 mL
3. A sample of KI solution is diluted 3.5 times its initial volume. If the original concentration was 1.25 M, what is its new concentration?
You can assign any value you like for V1. I will use 1.0 mL.(1.25 M) (1.0 mL) = X (3.5 mL)X = .357 M
25.0 grams of methanol (CH3OH) is dissolved in 250 grams of water. Calculate the freezing point of the solution.
25 g/32.05 g mol-1 = .780 mol methanolm = .780 mol methanol/ .250 kg solvent = 3.12mDtf = Kf x mDtf = (1.86 Ckg/mol)(3.12 mol/kg) = 5.8 CFP = 0 -5.8 C = -5.8 C
5. What is the freezing and boiling points of a solution of Ba(NO3)2 that contains 62.5 grams of solute in 1.0 kg of solvent?
62.5 g/ 261.3 g mol-1 = .239 molm = .239 mol/1.00 kg = .239 mDtf = Kf x m Dtf = (1.86 C Kg/mol)(.239 mol/kg) x 3 = 1.33 CFP = 0-1.33 = -1.33 CDon’t forget to multiply by 2 because of the ionic particles
Dtb = Kb x m Dtfb = (1.86 C Kg/mol)(.239 mol/kg) x 3= .36 CBP = 100 + .36 = 100.36 C
6. The freezing point of an aqueous sodium chloride solution is –20 C. Calculate the molality of the solution.
