Question

A 100 mL solution was made by adding 1.43 g of \(\mathrm{Na_2CO_3 \cdot xH_2O}\). The normality of the solution is 0.1 N. The value of x is ______.
(The atomic mass of Na is 23 g/mol)

Hint:

Equivalents of \(\mathrm{Na_2CO_3}\) = N x V(L); the equivalent weight of \(\mathrm{Na_2CO_3 \cdot xH_2O}\) is M/2, since the n-factor is 2.

Answer 1

Correct Answer: 10

Equivalents in solution = \(N \times V = 0.1 \times 0.1 = 0.01\) eq.
n-factor of \(\mathrm{Na_2CO_3}\) (as a base, \(\mathrm{CO_3^{2-}}\) takes 2 H+) = 2, so moles = 0.005.
Molar mass = \(\frac{1.43}{0.005} = 286\) g/mol.
Molar mass of \(\mathrm{Na_2CO_3}\) = 2(23) + 12 + 48 = 106, so 106 + 18x = 286, giving 18x = 180, x = 10.
Answer: 10