Amount of heat liberated when 0.49 g of sulphuric acid is neutralized by NaOH is -57.3 kJ -5.73 kJ 0.573 kJ -0.573 kJ

Dear Student 

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.
When one mole of $H_{2}S{O}_4 reacts with NaOH$ will give the heat of neutralization as -57.3 kJ
So, when 0.49 g of $H_{2}S{O}_4 or 0.01 mol$ will give the heat of neutralization as -0.573 kJ 

So, the correct option is (4) 

Regards