# The amount of $H_2SO_4$ present in 1200ml of 0.1m solution is

$(a)\;11.76g\qquad(b)\;12.76g\qquad(c)\;13.76g\qquad(d)\;23.52g$

$\large\frac{w}{m}$$\times 100=M\times V_{m2} \large\frac{w}{98}$$\times 1000=1200\times 0.1$
$\therefore wH_2SO_4=11.76g$
Hence (a) is the correct answer.