Question

10 g of a crystalline metal sulphate salt when healed generates approximately 6.4 g of an anhydrous salt of the same metal. The molecular weight of the anhydrous salt is 160 g The number of water molecules present in the crystal is:

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

The correct option is D 5

Given,

Let the metal sulphate be $MS{O}_4$

$10g$ of $MS{O}_4$ on heating gives $6.4g$ of anyhdrous salt of it

i.e. $MS{O}_4.x{H}_{2}O$

Molecular weight=$160g$

Now, First we calculate no. of moles of anhydrous salt in $6.4g$

it is known by data that

$10g$ salt gave $6.4g$ anhydrous salt

$\Rightarrow$ No. of moles of salt in the final product

$=\frac{6.4g}{160g/mol}=0.04$ moles $⟶1$

Now, Loss of water=$10g-6.4g=3.6g$

$\therefore$ Number of moles of water lost =$\frac{3.6g}{18g/mol}=0.2$ moles

$[\because H_{2}O=18g/mol]$

Now, here the ratio of water to metal sulphate in

$MS{O}_4.x{H}_{2}O$ is equal to

$\frac{No.ofmolesofwater}{No.ofmolesofsalt}=\frac{0.2}{0.04}=5$ $[\therefore 1$ & $2]$

$\therefore$ Number of water molecules=$5$