Question

1.245 g sample of $CuSO_{4}.xH_{2}O$ dissolved in water and excess of $H_{2}S$ was passed. The filtrate contains liberated $H_{2}SO_{4}$ which requires 20 ml N/2NaOH solutions for neutralization. Calculate the value of x?

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (D)

$M _{\text {eq. }}$ of $CuSO _{4} \times xH _{2} O = M _{ eq .} H _{2} SO _{4}= M _{\text {eq }} NaOH$

$M _{\text {eq. }} CuSO _{4}= M _{\text {eq. }} NaOH$

$\frac{10^{3} \times \text { wt. }}{\text { Mol. mass }} \times$ v.f. $= N \times V$

$\frac{1.245}{ M } \times 2 \times 10^{3}=\frac{4}{20} \times 2$

$M =1.245 \times 2 \times 10^{3} \times 10$

$M =249$

Molecular weight of $CuSO _{4} \cdot xH _{2} O =63.5+32+64+18 x$

$=159.5+18 x$

$249=159.5+18 x$

So, $x =5$.