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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
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1292 Upvotes
Verified Answer
The correct answer is:
5
$$
\begin{aligned}
&\text {Mass of } \mathrm{H}_{2} \mathrm{O}=\frac{36}{100} \times 250=90 \mathrm{g}\\
&\text { Moles of } \mathrm{H}_{2} \mathrm{O}=\frac{90}{18}=5 \mathrm{mol}
\end{aligned}
$$
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.
\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.
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