Question

$1.61$ gm of $N{a}_{2}S{O}_4.10H_{2}O$ contains same number of oxygen atoms as present in $x$ gm of $H_{2}S{O}_{4.}$ The value of $x$ is:

• A. $1.28$
• B. $1.38$
• C. $1.58$
• D. $1.78$

Answer 1

Answer: (D)

$1.78$

The molar masses of $N{a}_{2}S{O}_4.10H_{2}O$ and $H_{2}S{O}_4$ are $322.2$ g/mol and $98$ g/mol respectively.
One molecule of $N{a}_{2}S{O}_4.10H_{2}O$ contains $14$ oxygen atoms. One molecule of $H_{2}S{O}_4$ contains $4$ oxygen atoms.

Hence, $\frac{massofN{a}_{2}S{O}_4.10H_{2}O}{molarmassofN{a}_{2}S{O}_4.10H_{2}O}\times 14\times N_A=\frac{massof{H}_{2}S{O}_4}{molarmassof{H}_{2}S{O}_4}\times 4\times N_A$

Substituting values in the above expression, we get
$\frac{1.61}{322.2}\times 14\times N_A=\frac{x}{98}\times 4\times N_A$

$x=1.78g$
Option D is correct.