A non-electrolytic and non-volatile solute is added to pure water, difference between freezing point and boiling point is now 105°C. Calculate mass of solute present in 500 g of solvent. (Given molar mass of solute = 120 g/mol, kb = 0.512 °Cm–1, kf =1.86 °Cm–1)

To solve the problem of determining the mass of a non-electrolytic and non-volatile solute added to pure water, we can utilize the concepts of freezing point depression and boiling point elevation. The difference between the boiling point and freezing point of water after the addition of the solute has been given as 105°C, which will guide our calculations.

Understanding the Relationships

When a solute is added to a solvent, it affects the boiling point and freezing point of the solution. We can express these effects with the following formulas:

• Boiling Point Elevation: ΔT_b = k_b * m
• Freezing Point Depression: ΔT_f = k_f * m

Here, ΔT_b and ΔT_f are the changes in boiling and freezing points, respectively, k_b and k_f are the ebullioscopic and cryoscopic constants of the solvent (water in this case), and m is the molality of the solution.

Combining Freezing and Boiling Point Changes

In this scenario, we know the total change in temperature due to the addition of the solute is 105°C. We can express this as:

ΔT_b + ΔT_f = 105°C

Calculating Individual Changes

Using the formulas for boiling point elevation and freezing point depression, we can substitute:

k_b * m + k_f * m = 105°C

Given that:

• k_b = 0.512 °C/m
• k_f = 1.86 °C/m

This leads to:

m (k_b + k_f) = 105°C

m (0.512 + 1.86) = 105°C

m (2.372) = 105°C

Finding the Molality

Now we can solve for molality (m):

m = 105°C / 2.372 = 44.24 mol/kg

Calculating the Number of Moles

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we have 500 g of water (which is 0.5 kg), we can find the number of moles of solute (n) using:

m = n / mass of solvent (kg)

Thus,

44.24 = n / 0.5

n = 44.24 * 0.5 = 22.12 moles

Calculating the Mass of the Solute

To find the mass of the solute, we can use the molar mass of the solute:

Mass of solute = n * molar mass

Given that the molar mass of the solute is 120 g/mol:

Mass of solute = 22.12 moles * 120 g/mol = 2654.4 g

Final Result

Therefore, the mass of the solute present in 500 g of solvent is approximately 2654.4 grams.