Question

The mass of $1\times {10}^{22}$ molecules of blue vitriol $(CuS{O}_4\cdot x{H}_{2}O)$ is $4.144$g. The value of $x$ will be:

• A. 5
• B. 2
• C. 1
• D. 6

Answer 1

The correct option is A 5

Molecular weight of $CuS{O}_4=159.6g$

$159.6g$ contains $6.023\times {10}^{23}$ molecules of $CuS{O}_4$

$\therefore 1\times {10}^{22}$ molecules= $\frac{1\times {10}^{22}\times 159.6}{6.023\times {10}^{23}}$

$=2.649g$ of $CuS{O}_4$

Given weight of blue vitrol= $4.144g$

$\therefore$ Weight of water= $4.144-2.649$

$=1.495g=\frac{1.495}{18}$

$=0.083$ moles

Mass of $1\times {10}^{22}$ molecules of $H_{2}O$

$=\frac{0.083\times 6.023\times {10}^{23}}{1\times {10}^{22}}=5$

$\therefore x=5$