`0.06` mole of `KNO_(3)` is added to `100cm^(3)` of water at `298 K`. The enthalpy of `KNO_(3)(aq)` solution is `35.8 kJ mol^(-1)`. After the solute is dissolved, the temerature of the solution will be

Options

A. 293 K
B. 298 K
C. 301 K
D. 304 K

Answer

A

Explanation

Enthalpy of solution of `KNO_(3)` is endothermic reaction
`KNO_(3) (s) + H_(2)O(l) rarr KNO_(3)(aq), Delta H = +35.8` kJ/mol
Total heat absorbed Q = 35.8 `xx` 0.06 kJ
Specific heat of water `C_(v)= 4.18 xx 10^(-3)kJ//g Q= m.C_(v). Delta T`
`35.8 xx 0.06 =4.18 xx 10^(-3) kJ//g K xx 100g xx Delta T`
[Mass of `100cm^(3)` (100mL) of water = 100g]
`Delta T= (35.8 xx 0.06kJ)/(4.18 xx 10^(-3) xx 100kJ//K) = 5.13K`
Final temperature of the solution `=298K- 5.13 K ~~ 293K`