Question

0.70 g of a sample of Na₂CO₃.xH₂O were dissolved in water and the volume was made to 100 ml; 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralization. Calculate the value of x.

[At. wt. Na = 23, C = 12, O = 16 and H = 1]

Answer 1

100 ml of sol. contains 0.7 g of Na₂CO₃

1000 ml of sol. contains = \(\frac{0.7 \times 1000}{100}\) = 7.0 g of Na₂CO₃.xH₂O

\(\therefore\) Strength of Na₂CO₃.xH₂O = 7.0 g/litre

Now 20 ml of Na₂CO₃.xH₂O sol. = 19.8 ml of N/10 HCI

\(\therefore\) Normality of Na₂CO₃.xH₂O sol. = \(\frac{19.8}{20 \times 10} = \frac{19.8}{200} N\)

\(\therefore\) Eq. wt. of Na₂CO₃.xH₂O = \(\frac{7.0 \times 200}{19.8} = 70.70\)

But eq. wt. of Na₂CO₃.xH₂O = \(\frac{Mol. wt.}2 \)

\(= \frac{2 \times 23 + 12 + 3 \times 6 + x \times 18}2\)

\(= 53 + 9x\)

But the calculated eq. wt. = 70.70

\(\therefore\) 53 + 9x = 70.70

9x = 70.70 - 53

x = 1.96

\(\approx\) 2