Calculate the molality of a $1L$ soluttion of $93\%H_{2}S{O}_4$ (w/V) . The density of the solution is $1.84gm{L}^{-1}$

The correct option is D $10.428m$
$93\%w/V$ means that :
The solution is $1000$ mL, containing $930$ g of $H_{2}S{O}_4.$
The weight of the solution will be
$=\text{Density}\times \text{Volume}=(1.84\times 1000)g=1840g$

The weight of the solvent $\left(H_{2}O\right)$ :
$\text{Weight of solution - Weight of solute}=(1840-930)g=910g$
$\therefore \text{Molality}=\frac{\text{moles of}{H}_{2}S{O}_4}{\text{weight of}{H}_{2}O\left(g\right)}\times 1000$
$=\frac{930/98}{910}\times 1000=10.428m.$