Question

100ml 0.1M HCI is mixed with 100 ml 0.1M NaOH. Identify the correct statement (s)

• A. $\Delta H=-0.137k.Cal$
• B. $\Delta H=-0.573kJ$
• C. $\Delta T={0.685}^{\circ }C$
• D. $\Delta T={2.865}^{\circ }C$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\Delta H=-0.137k.Cal$
B $\Delta H=-0.573kJ$
C $\Delta T={0.685}^{\circ }C$

100ml 0.1 M HCI + 100 M1 0.1m NaOH $\Rightarrow$
10 mili equivalents of HCI + 10 mili equivalents of NaOH
$\therefore \Delta H=10\times {10}^{-3}\times 13.7k.cal=0.137k.cal$
$\Delta H=10\times {10}^{-3}\times 57.3kJ=0.573kJ$
$Q=ms\Delta T\therefore \Delta T=\frac{Q}{ms}=\frac{0.137\times {10}^{3}Cal}{200\left(gms\right)\times 1\left(cal/gm\right)}$
$={0.685}^{\circ }C$