Question

The crystalline solid $MS{O_4}.n{H_2}O$ of molar mass of $250gmo{l^{ - 1}}$, the percentage of anhydrous salt is 64 by weight. The value of n is:
A. 2
B. 3
C. 5
D. 7

Answer 1

Hint: We have to calculate mass of water using the molar mass of the given crystalline solid and remaining mass percentage of water. From the calculated mass of water, we can calculate moles of water using the mass of water and molar mass of water.

Complete step by step answer:
Given data contains,
Molar mass of $MS{O_4}.n{H_2}O$ is $250gmo{l^{ - 1}}$.
Mass percentage of anhydrous salts is $64\% $.
We can calculate the mass percentage of water by subtracting the mass percentage of anhydrous salt from 100.
Mass percentage of water = $100 - {\text{Mass of anhydrous salt}}$
Mass percentage of water = $100 - 64$
Mass percentage of water = $36\% $
The mass percentage of water is nothing but the remaining mass of water that is present in the solution, so the remaining mass of water is $36g$.
Now lets us calculate the total mass of water.
We can calculate the total mass of water by multiplying the remaining mass of water with the molar mass of crystalline solid. The product is divided by 100 to get total mass of water.
Mass of water = $\dfrac{{36g}}{{100}} \times 250gmo{l^{ - 1}}$
Mass of water = $90g$
The total mass of water is $90g$.
We can calculate moles of water by dividing the total mass of water to the molar mass of water.
Molar mass of water is $18g/mol$.
Moles of water = $\dfrac{{{\text{Total mass of water}}}}{{{\text{Molar mass of water}}}}$
Moles of water = $\dfrac{{90g}}{{18gmo{l^{ - 1}}}}$
$\Rightarrow$ Moles of water = $5mol$
The value of n in $MS{O_4}.n{H_2}O$ is 5. The complete formula of the crystalline solid could be written as $MS{O_4}.5{H_2}O$.

So, the correct answer is Option C .

Note:
We can also calculate the mass of a substance using mass percentage. An example is given below.
Example: The mass of sodium chloride present in $35g$ of ${\text{3}}{\text{.5\% }}$ solution has to be calculated.
Mass percentage of the solution = $3.5\% $
Mass of solution = $35.0g$
The mass of sodium chloride is given as,
${\text{Mass percentage}} = \dfrac{{{\text{Grams of sodium chloride}}}}{{{\text{Grams of solution}}}} \times 100\% $
$3.5\% = \dfrac{{{\text{Grams of sodium chloride}}}}{{35.0g}} \times 100\% $
Mass percentage = $1.23g$
The mass of sodium chloride present in $35g$ of ${\text{3}}{\text{.5\% }}$ solution is $1.23g$.