Question

The amount of energy released when 20 ml of 0.5 M NaOH are mixed with 100 ml of 0.1 M HCl is x kJ. The heat of neutralisation (in kJ $mo{l}^{-1}$) is :

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

Answer 1

The correct option is C +100 x

$NaOH$ $HCl$
20 mL 100 mL
0.5 m 0.1 m
= 10 m moles = 10 m moles
$\Delta H_N=-571ks/mole$
1 mole $\to$ -57.1
$10\times {10}^{-3}moles\to -57.1\times {10}^{-3}\times 10=\frac{57.1}{100}$
$\frac{57.1}{100}=x$
$57.1=x.100$