Answer to Question #315902 in General Chemistry for Aria

Question

A solution of 30 percent ethanol (C2 H3 OH) by weight has a density of 0.96 g/mL at 25 percent C. Find its molality and molarity?

Answer 1

X(C₂H₅OH) = 30 %;

M(C₂H₅OH) = 46 g/mol;

d(solution) = 0.96 g/mL;

T = 25 C = 298 K;

If V(solution) = 100 mL (0.1 L), then:

m(solution) = V(solution) * d(solution) = 100 * 0.96 = 96 g;

m(C₂H₅OH) = m(solution) * X(C₂H₅OH)/100 % = 96* 30/100 = 28.8 g

m(solvent) = m(solution) - m(C₂H₅OH) = 96 - 28.8 = 67.2 g = 0.0672 kg;

n(C₂H₅OH) = m(C₂H₅OH)/M(C₂H₅OH) = 28.8/46 = 0.626 mol;

c(molarity) = n(C₂H₅OH)/V(solution) = 0.626/0.1 = 6.26 M;

m(molality) = n(C₂H₅OH)/m(solvent) = 0.626/0.0672 = 9.32 mol/kg.

Answer: c(molarity) = 6.26 M;

m(molality) = 9.32 mol/kg.