25 mL of a 2 N $HCl$ solution , 50 mL of a 4 N $HN{O}_3$ solution and $x$ mL of a 5 M $H_{2}S{O}_4$ solution are mixed together and the total volume is made upto 1 litre with water. 50 mL of this acid mixture exactly neutralizes 25 mL of a 1 N $N{a}_{2}C{O}_3$ solution. The value of $x$ is:

The correct option is A 25 mL
According to the question :
$HCl+HN{O}_3+H_{2}S{O}_4$ are mixed together and total volume of solution is 1 L
Hence Milliequivalent of $HCl$ + Milliequivalent of $HN{O}_3$ + Milliequivalent of $H_{2}S{O}_4$ = Milliequivalent of acid
$\Rightarrow$ $N_{HCl}{V}_{HCl}$ +$N_{HN{O}_3}{V}_{HN{O}_3}$ + $N_{H_{2}S{O}_4}{V}_{H_{2}S{O}_4}=N_{\text{acid mixture}}{V}_{\text{acid mixture}}$

$\Rightarrow [2\times \frac{25}{1000}]+[4\times \frac{50}{1000}]+[3\times 2\times \frac{x}{1000}]=N_{\text{acid mixture}}\times 1$

$N_{\text{acid mixture}}=\frac{50+200+10x}{1000}N$
$N_{\text{acid mixture}}=\frac{250+10x}{1000}N$

Milliequivalent of mixture = Milliequivalent of $N{a}_{2}C{O}_3$

$N_{\text{mixture}}\times 50$ = $N_{N{a}_{2}C{O}_3}\times V_{N{a}_{2}C{O}_3}$

$\Rightarrow \frac{250+10x}{1000}\times \frac{50}{1000}$ = $1\times \frac{25}{1000}$
$\Rightarrow \frac{50}{1000}(250+10x)=25$
$x=25\text{mL}$