Question

Hydrated sulphate of a divalent metal of atomic weight 65.4 loses 35.85% of its weight on dehydration. Find the number of molecules of water of crystallization in the formula of hydrated salt.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{The}\mathrm{formula}\mathrm{of}\mathrm{the}\mathrm{salt}\mathrm{will}\mathrm{be}{\mathrm{MSO}}_4.{\mathrm{nH}}_2\mathrm{O} \\ \mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{salt}\mathrm{will}\mathrm{be}=65.4+32+4\times 16+\mathrm{n}\times 18=161.4+18\mathrm{n} \\ \mathrm{On}\mathrm{dehyration},\mathrm{all}{\mathrm{H}}_2\mathrm{O}\mathrm{molecules}\mathrm{will}\mathrm{evaporate}\mathrm{and}\mathrm{the}\mathrm{formula}\mathrm{of}\mathrm{the}\mathrm{salt}\mathrm{becomes}{\mathrm{MSO}}_4\mathrm{having}\mathrm{molar}\mathrm{mass}161.4 \\ \mathrm{Therefore},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that} \\ 35.85\%\mathrm{of}(161.4+18\mathrm{n})=161.4 \\ \mathrm{or},\frac{35.85}{100}\times (161.4+18\mathrm{n})=161.4 \\ \mathrm{or},161.4+18\mathrm{n}=\frac{161.4\times 100}{35.85} \\ \mathrm{or},161.4+18\mathrm{n}=450.20 \\ \mathrm{or}18\mathrm{n}=450.20-161.4=288.80 \\ \mathrm{or},\mathrm{n}=\frac{288.80}{18}\approx 16 \\ \mathrm{Hence},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{molecules}\mathrm{of}\mathrm{water}\mathrm{of}\mathrm{crystallisation}=16 \end{gathered}$$