Question

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 neutralisation (in kJ
mol-') is :-
(1) -100 x
(2) -50 x
(3) +100 x
(4) +50x​

Answer 1

Answer : the correct option is, (1) -100 x

Explanation :

First we have to calculate the moles of $NaOH$ and $HCl$.

$\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume}=(0.5mole/L)\times (0.02L)=0.01mole$

$\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.1mole/L)\times (0.1L)=0.01mole$

Now we have to calculate the heat of neutralization for 1 mole.

As, 0.01 mole releases heat = -X kJ

So, 1 mole release heat = $-\frac{X}{0.01}=-100XkJ/mole$

Therefore, the heat of neutralization will be, -100X kJ/mole