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

Options

A. 1.292 kJ
B. 2.292 kJ
C. 0.292 kJ
D. 22.9 kJ

Answer

B

Explanation

500 `cm^(3)` of 0.1 M HCI will give
`[(0.1)/(1000)xx500]` =0.05 moles of `H^(+)` ions
and 200 `cm^(3)` of 0.2 M NaOH will give
`[(0.2)/(1000)xx200]` =0.04 moles of `OH^(-)` ions 0.04 moles of `OH^(-)` ions can only react with 0.04 moles of `H^(+)` ions to form 0.04 moles of water.The amount of heat evolved during formation of 1 mole of `H_(2)O` is 57.32 KJ.
Heat evolved during formation of 0.04 moles of water is
`0.04xx57.32=2.2928 KJ`