Question

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1)A saturated solution of H2O2 in a water contains 30.0% by mass H2O2 and has a density of 1.11g/ml. Calculate the mole fraction, molarity & molality of this solution.

2) Calculate the molecular mass of an unknown volatile solute if 19.0g of it, dissolved in 500.0 g of pure water reduces the vapour pressure of the water from 23.80mmHg to 23.20 mmHg.

Answer 1

1)

saturated

m = 30% by mas of H2O

D = 1.11

assuem a basis of 100 g of solution

then

V = m/D = 100/1.11 = 90.0900ml = 0.09009 L

m = 30 g of H2O2

mol H2O2 = mass/MW = 30/34 = 0.88235

m = 70 g of water = 0.07 kg

mol water = 70/18 = 3.8888

mole fraction of H2O2 = mol of H2O2 / total mol = 0.88235 / (0.88235+3.8888) = 0.1849344

molarity = mol /L = 0.88235 /0.09009 = 9.794 M

molality = mol of solute / kg solvent = 0.88235 / 0.07 = 12.605 molal

2)

MW of unkonw

m = 19 g

m = 500 g owater = 0.5 kg water

mol water = mass/MW = 500/18 = 27.777

dP = 23.80 mmHg - 23.20 hg = 0.60 mm Hg

then

dP = xsolute*P°solvent

0.60 = xsolute*23.80

xsolute = 0.6/23.80 = 0.025210

x solute= mol solute / (mol S + mol W)

0.025210 = mol S /(mol S + 27.777)

0.025210*molS + 27.777*0.025210 = mol S

27.777*0.025210 = (1-0.025210)*S

S = 27.777*0.025210 / ((1-0.025210)) = 0.71836 mol of S

MW = mass/mol = 19/0.71836 = 26.44913g/mol