Question

1.245 g sample of $CuS{O}_4.x{H}_{2}O$ dissolved in water and excess of $H_{2}S$ was passed. The filtrate contains liberated $H_{2}S{O}_4$ 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 $CuS{O}_4\times x{H}_{2}O=M_{eq.}{H}_{2}S{O}_4=M_{\text{eq }}NaOH$

$M_{\text{eq. }}CuS{O}_4=M_{\text{eq. }}NaOH$

$\frac{10^3\times \text{ wt. }}{\text{ Mol. mass }}\times$ 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 $CuS{O}_4\cdot x{H}_{2}O=63.5+32+64+18x$

$=159.5+18x$

$249=159.5+18x$

So, $x=5$.