calcualate the amount of heat evolved when `500 cm^(3)` of 0.1M HCl is mixed with `200 cm^(3)` of 0.2 M NaOH.
A. 57.3 kJ
B. 2.865 kJ
C. 2.292 kJ
D. 0.573 kJ
Correct Answer - c
`HCl+NaOHto NaCl+H_(2)O`
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 ecolved = 57. 3 kJ
to neutralised 0.04 moles of NaOH by 0.04 molw of NaOH, heat evolved
`57.3 xx 0.04`
= 2.292 kJ
Correct Answer - A
Number of moles of `NaOH=(MV)/(1000)=(0.5xx20)/(1000)=0.01`
Number of moles of `HCl=(MV)/(1000)=(0.1xx100)/(1000)=0.01`
Heat of neutralization `=(-x)/(0.01)=-100x`