Some Basic Concepts of Chemistry - Result Question 56
55. The density of a $3 M$ sodium thiosulphate solution $\left(Na _2 S _2 O _3\right)$ is $1.25 g$ per $mL$. Calculate (i) the percentage by weight of sodium thiosulphate (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of $Na^{+}$and $S _2 O _3^{2-}$ ions.
(1983,5 M)
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Answer:
Correct Answer: 55. (i) 37.92, (ii) 0.065, (iii) 7.73m
Solution:
- (a) Let us consider $1.0 L$ solution for all the calculation.
(i) Weight of $1 L$ solution $=1250 g$
Weight of $Na _2 S _2 O _3=3 \times 158=474 g$
$\Rightarrow$ Weight percentage of $Na _2 S _2 O _3=\frac{474}{1250} \times 100=37.92$
(ii) Weight of $H _2 O$ in $1 L$ solution $=1250-474=776 g$
Mole fraction of $Na _2 S _2 O _3=\dfrac{3}{3+\frac{776}{18}}=0.065$
(iii) Molality of $Na^{+}=\frac{3 \times 2}{776} \times 100=7.73 m$
BETA
