Question

A salt mixture (1.0 g) contains 25 wt% of MgSO$_4$ and 75 wt% of M$_2$SO$_4$. Aqueous solution of this salt mixture on treating with excess BaCl$_2$ solution results in the precipitation of 1.49 g of BaSO$_4$. The atomic mass of M (in g mol$^{-1}$) (rounded off to two decimal places) is ________.

Hint:

Mass of BaSO$_4$ precipitate directly gives total sulfate ions, helping to find unknown cations through stoichiometry.

Answer 1

Correct Answer: 38.98

Step 1: Let moles of MgSO$_4$ = $n_1$, and M$_2$SO$_4$ = $n_2$. \[ 0.25 = \frac{n_1 \times 120.37}{1}, \quad 0.75 = n_2 (M_r(\text{M}_2\text{SO}_4)) = n_2(2M + 96.06) \] Total moles of sulfate ions = $n_1 + n_2$.
Step 2: BaSO$_4$ formed. 1.49 g BaSO$_4$ $\Rightarrow$ moles = $\frac{1.49}{233.39} = 6.38 \times 10^{-3}$ mol. So, \[ n_1 + n_2 = 6.38 \times 10^{-3} \] Step 3: Express $n_1$ and $n_2$ in terms of M. \[ n_1 = \frac{0.25}{120.37} = 2.076 \times 10^{-3} \] \[ n_2 = \frac{0.75}{2M + 96.06} \] Hence, \[ 2.076 \times 10^{-3} + \frac{0.75}{2M + 96.06} = 6.38 \times 10^{-3} \] \[ \frac{0.75}{2M + 96.06} = 4.304 \times 10^{-3} \Rightarrow 2M + 96.06 = 174.17 \] \[ M = 23.12 \] Step 4: Conclusion.
The atomic mass of M = 23.12 g mol$^{-1}$.