Question

a 5.0g sample of an ionic compound (MW=96.5g/mol) is placed in 50.0g of water at 25C and the solution is stired. When the salt is completely dissolved the temperature of the solution is 32.7C

a) if we assume that the specific heat of the solution is 1.0cal/g, calculate the *triangle-Hsoln* for this compound in Kcal/g and in Kcal/mol

Answer 1

Answer on Question #55940 - Chemistry - General chemistry

Question:

a 5.0g sample of an ionic compound (MW=96.5g/mol) is placed in 50.0g of water at 25°C and the solution is stirred. When the salt is completely dissolved the temperature of the solution is 32.7°C

a) if we assume that the specific heat of the solution is 1.0 cal/g, calculate the *triangle-Hsoln* for this compound in Kcal/g and in Kcal/mol

Answer:

$$Q = - \Delta H;$$ $$\Delta H_{soln} = - Q = - \frac{cm(T_2 - T_1)}{m_{sample}} = - \frac{1.0 \frac{cal}{g} \times (5.0 g + 50.0 g)(32.7^\circ C - 25^\circ C)}{5.0 g} = 84.7 \frac{cal}{g}$$ $$\approx 85 \frac{cal}{g}$$ $$\Delta H_{soln} = - Q = - \frac{cm(T_2 - T_1)}{n_{sample}} = - \frac{cmMW_{sample}(T_2 - T_1)}{n_{sample}}$$ $$= - \frac{1.0 \frac{cal}{g} \times (5.0 g + 50.0 g) \times 96.5 \frac{g}{mol} \times (32.7^\circ C - 25^\circ C)}{5.0 g} = 8173.55 \frac{cal}{mol}$$ $$\approx 8.2 \frac{kcal/mol}{mol}$$

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