Question

What would be the molality of a solution made by mixing equal volumes of $30.0\%$ by mass of $H_{2}S{O}_4$ (density $1.218gc{m}^{-3}$) and $70\%$ by mass of $H_{2}S{O}_4$ (density $1.610gc{m}^{-3}$)?

Answer 1

Let 100 mL of one solution be mixed with 100 mL of the other solution.
Mass of 100 mL of $30\%H_{2}S{O}_4=100\times 1.218=121.8g$

Mass of $H_{2}S{O}_4$ in 121.8 g of $30\%H_{2}S{O}_4$
$=\frac{30}{100}\times 121.8g=36.54g$

Mass of water $=(121.8-36.54)=85.26g$

Mass of 100 mL of $70\%H_{2}S{O}_4=100\times 1.61=161.0g$

Mass of $H_{2}S{O}_4$in 161.0 g of $70\%H_{2}S{O}_4$
$=\frac{70}{100}\times 161.0g=112.7g$

Mass of water $=(161.0-112.7)=48.30g$
Total $H_{2}S{O}_4$ in solution = 36.54 + 112.7 = 149.24 g

No. of moles of $H_{2}S{O}_4=\frac{149.24}{98}$

Total mass of water in solution = (85.26 + 48.30)
$=133.56g=\frac{133.56}{1000}kg$

Molality $=\frac{No.ofmolesof{H}_{2}S{O}_4}{Massofwaterinkg}=\frac{149.24}{98}\times \frac{1000}{133.56}$
$=11.4m$