Question

How many mL of sulphuric acid of density 1.84 g mL${}^{-1}$ containing 95.6 mass % of $H_{2}S{O}_4$ should be added to one litre of 40 mass % solution of $H_{2}S{O}_4$ of density 1.31 g mL${}^{-1}$ in order to prepare 50 mass % solution of sulphuric acid of density 1.40 g mL${}^{-1}$?

• A. 66.2 ml
• B. 55.2 ml
• C. 95.5 ml
• D. 105.2 ml

Answer 1

The correct option is C 95.5 ml

Let mass of solution be 100 gm
​Therefore amount of solute in it = 95.6 gm
Volume of solution = $\frac{mass}{density}$ = $\frac{100}{1.84}$ = 54.3478 ml
Volume = 1 L =1000 ml
Density = 1.31 g/ml
Mass of solution = 1310 g
Mass % of solute = 40 %

Let mass of solute in it be X g

Mass % = $\frac{massofsolute}{massofsolution}\times 100$

40 = $\frac{X}{1300}\times 100$

X = $40\times 1310$ = 524 gm

For solution to make 50 % mass percent and density 1.40 g/ml
Volume of solution =1400 ml
Amount of solute present should be,

Y gm = $\frac{50\times 1400}{100}$ = 700 g
​Thus
Amount of solute needed to be added to the 40 % to make it 50% by mass =700-524 = 176 g
Volume of H2SO4 of density 1.84 g/ml required to be added to 40 % = $\frac{mass}{density}$ = $\frac{176}{1.84}$ =95.6521 ml

so the answer is C