Question

Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving $4$ g of sulphur trioxide in a $100$ml sulphuric acid solution containing $80$ mass percent (w/w) of $H_{2}S{O}_4$ and having a density of $1.96g/ml$(molecular weight of $H_{2}S{O}_4=98$) Take reaction $S{O}_3+H_{2}O\to H_{2}S{O}_4$.

• A. $80.8\%$
• B. $84\%$
• C. $41.65\%$
• D. None of these

Answer 1

The correct option is A $80.8\%$

$S{O}_3+H_{2}O\to H_{2}S{O}_4$

100 ml sulphuric acid solution

density = $\frac{Mass}{volume}$

$1.96g/ml=\frac{Mass}{100}$

Mass $=1.96g$

Amount of water $=196\times 20$

$=39.2g$

Amount of $H_{2}S{O}_4=156.8g$

$S{O}_3+H_{2}O\to H_{2}S{O}_4$

Molar mass of $S{O}_3=32+16\times 3=80$

No. of mole = $49$

$80g=0.05mol$

0.05 mol of sulphuric acid formed

Now total amount of sulphuric acid

$=0.05\times 98+156.8$

$=4.9+156.8$

$=161.7$

total amount of water $=39.2-$ total water consume

$=39.2-0.05\times 18$

$=39.2-0.9=38.3$

Mass percent (w/w) $=\frac{161.7}{161.7+383}=\frac{161.7}{200}\times 100\%$

$=80.85\%$

$\approx 80.8\%$

Option A is correct.