Question

$20c{m}^3$ of $0.5M$ $NaOH$ requires $30c{m}^3$ of $HCl$ solution as titrant. Waht mass of $HCl$ has been consumed in the reaction.

• A. $0.365g$
• B. $.65g$
• C. $36.5g$
• D. None

Answer 1

Answer: (A)

$0.365g$

$M_1{V}_1=M_2{V}_2$

Where, $M_1$ and $M_2$ are the molaraties of $NaOH$ & $HCl$ respectively & $V_1$ and $V_2$ are their respective volumes.

$\therefore 0.5\times 20=M_2\times 30$

$\therefore M_2=\frac{0.5\times 20}{30}=0.333M$

$\therefore \frac{molesofHCl}{0.030}=0.333$

$\therefore$ Mass of $HCl$ $=0.333\times 0.030\times 36.5$

$\therefore$ Mass of $HCl$ =0.365g