Q15 / 155 · WBJEE 2019 · Chemistry

In the crystalline solid $M \mathrm{SO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O}$ of molar mass 250 g $mol^{-1}$ , the percentage of anhydrous salt is 64 by weight. The value of $n$ is

  • A
    2
  • B
    3
  • Correct answer
    C
    5
  • D
    7

Solution

$$ \begin{aligned} &\text {Mass of } \mathrm{H}_{2} \mathrm{O}=\frac{36}{100} \times 250=90 \mathrm{g}\\ &\text { Moles of } \mathrm{H}_{2} \mathrm{O}=\frac{90}{18}=5 \mathrm{mol} \end{aligned} $$ In the crystalline solid $M \mathrm{SO}_{4} \cdot n \mathrm{H}_{2} \mathrm{O},$ the value of $n$ is 5 Hence, option (c) is correct answer.

Hence, the correct answer is option: 'C'.

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