Question

How many $mL$ of a 0.25 N solution of $HCl$ are needed to completely neutralize 5 grams of crystalline $NaOH$ dissolved in water?

• A. 150
• B. 250
• C. 500
• D. 750

Answer 1

The correct option is C 500

$N_1{V}_1=N_2{V}_2$

Also, $N=\frac{Numberofgramequivalentsofsolute}{Volumeofsolution\left(in\right]L)}$

So, $N\times V\left(inL\right)=Numberofgramequivalentsofsolute$

For $NaOH$, $Equivalentweight=\frac{MolarMass}{n-factor}=\frac{40}{1}$

and Number of gram equivalents of solute $=\frac{Givenmass}{equivalentweight}=\frac{5\times 1}{40}=\frac{5}{40}$

So, $N_1{V}_1=N_2{V}_2$

$0.25\times V_1=\frac{5}{40}$

$V_1=\frac{5}{40\times 0.25}=0.5L=500mL$