Question

Calculate the molality of a solution made by mixing equal volumes of $30$% by weight of $H_2{SO}_4$ ($d=1.218g/mL$) and $70$% by weight of $H_2{SO}_4$ ($d=1.610g/mL$).

• A. 8.22
• B. 9.22
• C. 11.22
• D. 7.57

Answer 1

The correct option is C 11.22

Let we take $x$ litre of first

Solution of $H_2{SO}_4$ i.e $30\%$ by weight

Then we also have to take $x$ liter te of second solution of $H_2{SO}_4$ (i.e: $70\%$ by weight)

in first solution, weight of solution $=x\times$ density

$\begin{array}{c}=x\times 1.218\\ =1.218xkg\end{array}$

$\begin{array}{cc}(\text{density}& =1.218g/ml\\ & =1.218kg/l\end{array})$

Mass of $H_2$ SO ${}_4$ in this solution

$=30\%\text{of}1.218x$

$=\frac{30}{100}\times 1.218x$

$=0.3654\times kg$

Mass of solvent $=(1.218-0.3654)$

$x\times$

$=0.8526xkg$

in second solution,

weight of solution $=x\times$ density

$\begin{array}{c}=x\times 1.610\\ =1.610xkg\end{array}$

( density $=1.610g/ml$

$=1.610kg/l$

Mass of $H_2{SO}_4$ in this solution

$=70\%$ of $1.610x$

$\begin{array}{cc}=\frac{70}{100}x+160x=& 1.127xkg\\ \text{Mass of solvent}& =(1.610-1.127)x\\ & =0.483xkg\end{array}$

After adding both the solution,

Mass of of solvent $=(0.483+0.8526)x$

$=1.3356xkg$

$\begin{array}{cc}\text{Mass of}{H}_2{SO}_4& =(0.3654+1.127{)}^x\\ & =1.4924\times kg\end{array}$

$=1492.4xg$

Moles of $H_{2}S{O}_4=\frac{1429\cdot 4x}{98}$

$=15\cdot 23x$

Molality $=\frac{Moleof{H}_{2}S{O}_4}{massofsolventkg}$

$=\frac{15.23x}{1.3356x}$

$=11\cdot 22m$