Question

How many gm of $H_2{SO}_4$ of density 1.84$/$ containing 95.6 mass % of $H_2{SO}_4$ should be added to 1 L of 40 mass % solution of $H_2{SO}_4$ of density 1.31$g/l$ in order to prepare 50 mass % solution of sulphuric acid of density 1.4 $g/l$

Answer 1

Assume the mass of solution= $100gm$

Amount of $H_{2}S{O}_4=95.6gm$

Volume of solution= $\frac{mass}{density}=\frac{100}{1.84}=54.34ml$

Given that volume= $1lit=1000ml$

Density= $1.31gm/ml$

Mass of solution= Density $\times$ Volume=$1310gm$

Mass of % of solute $\left(H_{2}S{O}_4\right)=50$%

Mass % of solute= $\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$\Rightarrow 40=\frac{M}{1310}\times 100\Rightarrow M=524gm$

To make $50$% mass percent and density $1.40g/ml$

Volume of solution= $1400ml$

Mass % of solute= $\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$\Rightarrow M=700gm$

Amount of $H_{2}S{O}_4$ needed to add to $40$% to make it $50$% by mass= $700-524=176gm$

Volume required= $\frac{176}{1.84}=95.65ml$