Question

82 mg of solute is mixed into a 0.35 L solution. At a temperature of 37°C and an osmotic
pressure of 3.2 x 10-3 atm, calculate the molar mass of the solute.

Answer 1

Solution.

m(solute) = 82 mg = 0.082 g;

V(solution) = 0.35 L;

T = 310 K;

$\pi$ =3.2 x 10^-3 atm =3.2 x 10^-3 x 101 325 Pa = 324.24 Pa;

M(solute) - ?;

$\pi$ = c$\sdot$ R$\sdot$ T; c - molar concentration;$\implies$

c = $\pi$ /(R⋅ T); c =324.24 Pa/(8.31J/(mol$\sdot$ K)$\sdot$310 K) = 0.126mol/L;

c = $\nu$( solute)/V(solution); $\implies$ $\nu$ (solute) = c$\sdot$ V(solution);

$\nu$ (solute) = 0.126mol/L$\sdot$ 0.35 L = 0.0441 mol;

$\nu$ (solute) = m/M; $\implies$M = m/$\nu$(solute);

M =0.082 g/0.0441 mol = 1.86 g/mol$\approx$2 g/mol;

Answer: M = 2 g/mol.