Question

When a 7.00-g sample of RbBr is dissolved in water in a calorimeter that has a total heat capacity of 1.931 kJ·K–1, the temperature decreases by 0.480 K. Calculate the molar heat of solution of RbBr.

Answer 1

Suppose q is the quantity of heat transferred to water in the calorimeter by dissolving 7.00 g sample of Rb Br , then quantity of heat absorbed by the calorimeter ,

q = Cv x T ; [ heat capacity of calorimeter Cv is given as 1.931kJ K^-1

... = 1.931 x ( - 0.480 ) k J ,[ since there is a decrease in temperature T would have a negative sign ]

... = - 0.92688 k J

Quantity of heat from the dissolution of RbBr will have the same magnitude but opposite sign ,because heat exchange in the system by water and the calorimeter should be equal

so ,..............q = + 9.2688 x 10 ^{- 1} kJ

( Note - here positive sign indicates the endothermic nature of the heat of solution.)

Thus heat absorbed by dissolving 7.00 gms of RbBr = 9.2688 x 10^-1 kJ

& --------------------------one mole or =164.38 gms --------- = ( 9.2688 x 10^-1 x 164.38 ) / 7.00

............................................................................. = 21.89 kJ / mole

Thus the molar heat of of solution of RbBr = 21.89 k J / mole