In the crystalline solid MSO_{4} cdot nH_{2}O | vedclass

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(C) The molar mass of the crystalline solid $MSO_{4} \cdot nH_{2}O$ is $250 \ g \ mol^{-1}$.Given that the percentage of anhydrous salt $(MSO_{4})$ is

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$5$

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(C) The molar mass of the crystalline solid $MSO_{4} \cdot nH_{2}O$ is $250 \ g \ mol^{-1}$.Given that the percentage of anhydrous salt $(MSO_{4})$ is $64\%$, the percentage of water $(H_{2}O)$ is $100\% - 64\% = 36\%$.Mass of $H_{2}O = \frac{36}{100} 	imes 250 = 90 \ g$.Moles of $H_{2}O = \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.Since the formula is $MSO_{4} \cdot nH_{2}O$, the value of $n$ corresponds to the number of moles of water, which is $5$.Hence, option $(C)$ is the correct answer.

In the crystalline solid $MSO_{4} \cdot nH_{2}O$ 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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