In: Chemistry
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