Question

The density of 3M solution of $N{a}_2{S}_2{O}_3$ is $1.25$ g mL${}^{-1}$. Calculate the $\%$ by weight of $N{a}_2{S}_2{O}_3$ and molality of $N{a}^{+}$ ion

• A. 62.08 $\%$, 7.73 M
• B. 37.92 $\%$, 7.73 M
• C. 42.73 $\%$, 3.86 M
• D. 58.58 $\%$, 3.86 M

Answer 1

The correct option is B 37.92 $\%$, 7.73 M

Given $N{a}_2{S}_2{O}_3$ has molarity = 3 mol L${}^{-1}$. This implies that 3 mol of $N{a}_2{S}_2{O}_3$ are present in 1 L of solution.
$\therefore$ Weight of $N{a}_2{S}_2{O}_3=Molarmass\times Mole$
$=3\times 158=474$ g
and Volume of solution $=$ 1 litre $=1000mL$
Also density of solution is 1.25 gml${}^{-1}$
Hence weight of solution $=1000\times 1.25=1250$ g
Weight of solvent i.e water $=1250-474=776$ g
$\%$ by weight of $N{a}_2{S}_2{O}_3=\frac{\text{Weight of}N{a}_2{S}_2{O}_3}{\text{Weight of}N{a}_2{S}_2{O}_3+\text{Weight of}{H}_{2}O}$
$\Rightarrow \frac{474}{1250}\times 100=37.92$
3 mol of $N{a}_2{S}_2{O}_3$ gives 6 mol of $N{a}^{+}$ and 3 mol of $S_2{O}_3^{2-}$
Molality of $N{a}^{+}=\frac{\text{Mole of}N{a}^{+}}{\text{Weight of water (g)}}\times 1000=\frac{6\times 1000}{776}=7.732m$