# Question 6ec6f

Apr 18, 2017

The mass % or $\frac{w}{w}$ % of the ${H}_{2} S {O}_{4}$ solution is 29%#
This means $100$ g of the solution contains $29$ g or $\frac{29 g}{98 g \text{/} m o l} = \frac{29}{98} m o l$ ${H}_{2} S {O}_{4}$.
Now let the density of the solution be $d$ $g \text{/} m L$

So volume of 100 g ${H}_{2} S {O}_{4}$ solution is $\frac{100}{d}$ mL or
$\frac{0.1}{d}$ L

So molarity of the solution is $\frac{\frac{29}{98}}{\frac{0.1}{d}}$ M

So by the problem

$\frac{\frac{29}{98}}{\frac{0.1}{d}} = 3.6$

$\implies d = \frac{3.6 \times 9.8}{29} \approx 1.2$ $g \text{/} m L$