Some Basic Concepts of Chemistry 1 Question 35
55. The density of a $3 \mathrm{M}$ sodium thiosulphate solution $\left(\mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}{3}\right)$ is $1.25 \mathrm{~g}$ per $\mathrm{mL}$. Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $\mathrm{Na}^{+}$and $\mathrm{S}{2} \mathrm{O}_{3}^{2-}$ ions.
$(1983,5 \mathrm{M})$
Show Answer
Answer:
Correct Answer: 55. (i) 37.92 , (ii) 0.065 , (iii) $7.73 \mathrm{~m}$ 56. (a) 0.6 , (b) 24
$\begin{array}{lll}\text { 58. (i) } 0.0179 \mathrm{~g} \text {, (ii) } 10.6 % & \text { 59. }(0.437) & \text { 61. } 20 %\end{array}$
Solution:
- (a) Let us consider $1.0 \mathrm{~L}$ solution for all the calculation.
(i) Weight of $1 \mathrm{~L}$ solution $=1250 \mathrm{~g}$
Weight of $\mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}_{3}=3 \times 158=474 \mathrm{~g}$
$\Rightarrow$ Weight percentage of $\mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}_{3}=\frac{474}{1250} \times 100=37.92$
(ii) Weight of $\mathrm{H}_{2} \mathrm{O}$ in $1 \mathrm{~L}$ solution $=1250-474=776 \mathrm{~g}$
Mole fraction of $\mathrm{Na}{2} \mathrm{~S}{2} \mathrm{O}_{3}=\frac{3}{3+\frac{776}{18}}=0.065$
(iii) Molality of $\mathrm{Na}^{+}=\frac{3 \times 2}{776} \times 100=7.73 \mathrm{~m}$
