The amount of heat evolved when `500 cm^(3) 0.1 M HCl` is mixed with `200 cm^(3)` of `0.2 M NaOH` is

Options

A. 57.3 kJ
B. 2.865 kJ
C. 2.292 kJ
D. 0.573 kJ

Answer

C

Explanation

`HCl + NaOH rarr NaCl+H_2O`
At t = 0
Number of moles
`= (500xx0.1)/(1000)(200xx0.2)/(1000)`
`=0.05 = 0.04`
`:.` During neutralisation of 1 mole of NaOH by 1 mole of HCl, heat evolved = 57.3 kJ
`:.` To neutralized 0.04, moles of NaOH by 0.04 mole of NaOH, heat evolved
`=57.3 xx0.4`
`= 2.292 kJ`