Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Percent by Mass - Page 110: 60

$V=631.31ml$

$\omega =18\%=0.18$ $\rho = 1.1\frac{g}{ml}$ $m(solute)=125g$---$V=?$ $m(solute)=\omega \times m(solution)=\omega \times \rho \times V \implies V=\frac{m(solute)}{\omega \times \rho}=\frac{125g}{0.18\times 1.1\frac{g}{ml}}=631.31ml$ Hence, $631.31ml$ of this solution contains $125g$ of $(NH_4)_2SO_4$.