Some Basic Concepts of Chemistry - Result Question 56

Answer:

Correct Answer: 55. (i) 37.92, (ii) 0.065, (iii) 7.73m

Solution:

55. (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 $Na^{+}=\frac{3 \times 2}{776} \times 100=7.73 m$