Answer to Question #113977 in General Chemistry for Boluwatife Obasanjo

Question

The density of acetonitrile (CH3CN) is 0.786 g/mL and the density of methanol (CH3OH) is 0.791 g/mL. A solution is made by dissolving 15.7 mL of CH3OH in 92.3 mL of CH3CN. What is the mole fraction of the methano in this solution?

Answer 1

ρ(CH₃CN) = 0.786 g/mL;

ρ(CH₃OH) = 0.791 g/mL;

V(CH₃CN) = 92.3 mL;

V(CH₃OH) = 15.7 mL;

M(CH₃CN) = 41.05 g/mol;

M(CH₃OH) = 32.04 g/mol;

m = ρ×V;

m(CH₃CN) = 0.786×92.3 = 72.54 g;

m(CH₃OH) = 0.791×15.7 = 12.42 g;

n = m/M;

n(CH₃CN) = 72.54/41.05 = 1.76 moles;

n(CH₃OH) = 12.42/32.04 = 0.38 moles;

Mole fraction X(CH₃OH) is = (moles of CH₃OH)/(total moles);

X(CH₃OH) = 12.42/(72.54 + 12.42);

Answer: 0.14