The amount of heat released, when 20 mL of 0.5 M NaOH is mixed with 100 mL of 0.1 M HCl, is x kJ.The heat of neutralization (in kJ `mol^(-1)` ) is:

Options

A. `-100x`
B. `-50x`
C. `+100x`
D. `+50x`

Explanation

Number of moles of `NaOH = ( MV )/(1000) = ( 0. 5 xx 20)/( 1000) = 0. 01`
Number of moles of `HCl = ( MV)/(1000) = ( 0.1 xx 1000)/(1000) = 0. 01,` Heat of neutralization `= (-x)/(0.01) =- 100 x`