Question

0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. After addition of the solid, the solution temperature is raised to about 27.0 degrees Celsius. The substance is known to have a molar mass of about 56 g/mol. Calculate the enthlapy of solution in kJ/mol.

Answer 1

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

$Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J$

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = $\frac{0.9975 g}{56 g/mol}=0.01781 mol$

The enthlapy of solution :

$\frac{Q'}{moles}$

$=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol$

The enthlapy of solution is -55.23 kJ/mol.