Question

12) You are preparing 960.0 grams of an aqueous solution that contains that contains 32.00% by weight of Na2CO3. What mass of solute would you use and what mass of solvent would you use? [4 points) ider 13) What is the weight percent of a calcium chloride (CaCl2] solution that is made by dissolving 151.6 grams of CaCl in 648.4 grams of water? [4 points) 14) Calculate the new molarity of a 20.0-ml sample of 4.50 M MgCl2 solution that is diluted with enough water to make 500.0 mL of solution. [5 points)

Answer 1

12 )

Given : 1) Mass of solution = 960.0 g

2) Solute : sodium carbonate

3) Concentration of solution : 32.00 % w/w

We have , % w/w = [ Mass of solute / Mass of solution ] 100

$\therefore$ 32.00 = [ Mass of Sodium carbonate / 960.0 g ] 100

[ Mass of Sodium carbonate / 960.0 g ] = 32.00 / 100 = 0.320

Mass of Sodium carbonate = 0.320 ( 960.0 g ) = 307.2 g

Mass of solution = Mass of solute + Mass of solvent

$\therefore$ Mass of solvent = Mass of solution - Mass of solute = 960.0 g - 307.2 g = 652.8 g

ANSWER : We can use 307.2 g sodium carbonate and 652.8 g solvent to prepare 32.00 % w/w solution.

13)

Given : 1) Mass of solute ( CaCl 2) = 151.6 g

2) Mass of solvent (Water) = 648.4 g

Mass of solution = Mass of solute ( CaCl 2) + Mass of solvent (Water) = 151.6 g + 648.4 g = 800.0 g

% w/w = [Mass of solute / Mass of solution ] 100

$\therefore$ % w/w CaCl₂ = [151.6 g /800.0 g ] 100 = 18.95 %

ANSWER : 18.95 %

14 )

Given : Molairty of stock = 4.50 M

Volume of stock = 20.0 ml

Volume of dilute = 500.0 ml

Molarity of dilute = ?

We have dilution formula, M _stock$\times$ V _stock = M _dilute$\times$ V _dilute

M _dilute = M _stock$\times$ V _stock / V _dilute = 4.50 M $\times$ 20.0 ml / 500.0 ml = 0.180 M

ANSWER : Concentration of diluted MgCl ₂ solution is 0.180 M