Question

$150g$ of a solution $(d=1.2g/ml)$ contains $63g$ of $HN{O}_3$. What volume of this $HN{O}_3$ (in mL) solution will be required to react with a $NaOH$ solution, containing $24g$ $NaOH$?
Reaction: $HN{O}_3+NaOH\to NaN{O}_3+H_{2}O$

• A. $75$ ml
• B. $100$ ml
• C. $150$ ml
• D. $50$ ml

Answer 1

The correct option is A $75$ ml

Mole of $HN{O}_3=\frac{63}{63}=1;$

Volume of $HN{O}_3$ solution $=150/1.2=125$ mL

Molarity of $HN{O}_3=\frac{1}{125\times {10}^{-3}}=8M$;

Solution required for $NaOH=V$ L

Mole of $HN{O}_3$ required for $NaOH=8V$

Mole of $NaOH$ $=\frac{24}{40}=\frac{6}{10}$

$8V=\frac{6}{10}$

V $=0.075L=75$ mL.