Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Percent by Mass - Page 110: 59

$m(NH_4Cl)=85.05g$

$\omega =18\%=0.18$ $\rho = 1.05\frac{g}{ml}$ $V=450ml$---$m(solute)=?$ $m(solute)=\omega \times m(solution)=\omega \times \rho \times V = 0.18\times 1.05\frac{g}{ml}\times 450ml = 85.05g$ Hence, $85.05g$ of $NH_4Cl$ is required to prepare this solution.