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Ideal Gas Equation

200 ml of 1M ...

Question

200 ml of 1M HCl is mixed with 400 ml of 0.5M NaOH. The temperature rise in the calorimeter was found to be 4.0oC. Water equivalent of calorimeter is 25g and the specific heat of the solution is 1 cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is −13.5 kcal, then the percentage error in this experiment while calculating the heat of neutralization is

A

7.4

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B

3

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C

5.6

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D

4.5

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Solution

The correct option is A 7.4
Total volume $200 + 400 = 600 m l$
200 ml of 1 M acid $=$ 400 ml of 0.5 M $N a O H$
200 m. eq. of acid reacts with 200 m.eq. of base $= Δ H$
Heat of neutralization $= 5 \times Δ H$
Heat produced during neutralization $=$ heat taken up by calorimeter + solution
$= m_{1} s_{1} Δ T + m_{2} s_{2} Δ T$
$= (25 \times 1 \times 4) + (600 \times 1 \times 4) = 2500 c a l$
$∴$ Heat of neutralization $= - 5 \times 2500$ $= - 12500 c a l = - 12.5 k c a l$
% error $= (\frac{13.5 - 12.5}{13.5}) \times 100 = 7.4$

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