Question

# A solution is prepared by dissolving 20.2 mL of methanol (CH3OH) in 100.0 mL of water...

A solution is prepared by dissolving 20.2 mL of methanol (CH3OH) in 100.0 mL of water at 25 ∘C. The final volume of the solution is 118 mL. The densities of methanol and water at this temperature are 0.782 g/mL and 1.00 g/mL, respectively. For this solution, calculate each of the following.

A. Molarity

B. Molality

C. Percent by Mass

D. Mole fraction

V = 20.2 ml of methanol

V = 100 ml of water

V = 118 ml solution

Dmet = 0.782 g/ml

Dwat = 1 g/ml

a)

M = mol of methanol / V sol

mol = mass/MW = D*V/MW = (20.2*0.782)/(32) = 0.4936375 mol of methanol

M = 0.4936375/(0.118) = 4.18336

b)

molality = mol / kg solvent

kg solvent = 100 g = 0.1 kg

molality = (0.4936375)/(0.1) = 4.936375 molal

c)

% mass methanol = mass of methanol /total mass * 100 =

total mass = D1V1 + D2V2 = 20.2*0.782 + 100*1 = 115.7964 g

mass of methanol = D1V1 = 0.782*20.2 = 15.7964

% mass = 15.7964/115.7964 *100 = 13.6415%

d)

mole frac = mol methano / total mol

mol meth = 0.4936375

mol water = mass/MW = 100/18 = 5.55

total mol =  5.55+0.4936375 = 6.04363

mole frac = 0.4936375 / 6.04363 = 0.081678