Question

A solution of a specific gravity $1.6g{mL}^{-1}$ is $67\%$ by mass. The $\%$ by mass of the solution of the same acid, if it is diluted to the specific gravity $1.2g{mL}^{-1}$ is (as nearest integer) :

Answer 1

Assume $100$ mL of solution of $67\%$ by mass of specific gravity $1.6$ $g{mL}^{-1}$.
Mass of solution $=100\times 1.6=160g$.
Mass of solute $=\frac{67\times 160}{100}=107.2g$.
Now, consider $Xg{H}_{2}O$ is added to it.
Mass of new solution $=(160+X)g$ ...(i)
Also, $Xg{H}_{2}O$ means $X$ mL of $H_{2}O$ and, thus,
Volume of new solution $=$ $(100+X)$
Using specific gravity of the solution, the mass of new solution
$=(100+X)\times 1.2=(120+1.2X)g$ ...(ii)
By Eqs. (i) and (ii), we get
$160+X=120+1.2X$
$X$ $=200$ mL or $200$ g
$\%$ by mass of new solution
$=\frac{107.2}{160+200}\times 100=29.78\%$.
So, the answer is $30$.