Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Percent by Mass - Page 110: 58

$m((NH_4)_2SO_4)=148.5g$

$\omega =18\%=0.18$ $\rho = 1.1\frac{g}{ml}$ $V=750ml$---$m(solute)=?$ $m(solute)=\omega \times m(solution)=\omega \times \rho \times V = 0.18\times 1.1\frac{g}{ml}\times 750ml = 148.5g$ Hence, $148.5g$ of $(NH_4)_2SO_4$ is required to prepare this solution.