Given-
Mass of sample = 1.20g
Normality of HCl = 0.1 N
Volume of solution = 100 mL
To calculate normality of solution -
\(M_1 \times V_1 = N_2V_2\)
\(0.1 \times 40 = N_2 \times 20\)
\(N_2 = 0 .2N\)
\(\because\) 1L solution contain number of g equivalent = 0.2
0.1L of solution contains no of g equivalent = 0.1 x 0.2 = 0.02g equivalent
\(\because\) Number of g equivalent = \(\frac{\text{mass}}{\text{equivalent mass}}\)
equivalent mass of \(Na_2CO_3 = \frac{106}2 = 53 g\)
equivalent mass of \(K_2CO_3 = \frac{138}2 = 69 g\)
Let say mass of \(Na_2CO_3 = xg\)
\(\therefore\) mass of \(K_2CO_3 = (1.2 - x)g\)
so, we can write-
\(\frac{1.2 -x}{69} + \frac x{53} = 0.02\)
⇒ \(x = 0.59 g\)
weight of \(Na_2CO_3 \) in mixture is \(= 0.59g\)