Question

$200mL$ of $0.2MHCl$ is mixed with $300mL$ of $0.1MNaOH$. The molar heat of neutralization of this reaction is $-57.1kJ$. The increase in temperature in ${}^{\circ }C$ of the system on mixing is $x\times 10^{-2}$. The value of $x$ is - (Nearest integer)
[Given : Specific heat of water $=4.18J{g}^{-1}{K}^{-1}$ Density of water $=1.00gc{m}^{-3}$ ]
(Assume no volume change on mixing)

Answer 1

Answer: 82

$\Rightarrow$ Millimoles of $HCl=200\times 0.2=40$
$\Rightarrow$ Millimoles of $NaOH=300\times 0.1=30$
$\Rightarrow$ Heat released $=\left(\frac{30}{1000}\times 57.1\times 1000\right)=1713J$
$\Rightarrow$ Mass of solution $=500ml\times 1gm/ml=500gm$
$\Rightarrow \Delta T=\frac{q}{m\times C}=\frac{1713J}{500g\times 4.18\frac{J}{g-K}}=0.8196K$
$=81.96\times 10^{-2}K$