A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water. (Given: Freezing point of pure water = 273.15 K)

To calculate the freezing point of a 5% glucose solution in water, we can use the concept of molality and the freezing point depression equation.

First, let's calculate the molality of the 4% sucrose solution.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Given that the molar mass of sucrose (C12H22O11) is 342 g/mol, we can calculate the number of moles of sucrose in the solution.

4% solution means 4 g of sucrose in 100 g of solution. So, the mass of sucrose is 4 g.

Number of moles of sucrose = mass of sucrose / molar mass of sucrose = 4 g / 342 g/mol = 0.0117 mol

Now, let's calculate the molality of the sucrose solution.

Molality (m) = moles of solute / mass of solvent (in kg) = 0.0117 mol / 0.100 kg = 0.117 mol/kg

Next, we can use the freezing point depression equation to calculate the freezing point of the 5% glucose solution.

ΔTf = Kf * m

Where ΔTf is the freezing point depression, Kf is the cryoscopic constant (which is a property of the solvent), and m is the molality of the solution.

Given that the freezing point of pure water is 273.15 K and the freezing point of the 4% sucrose solution is 271.15 K, we can calculate the value of Kf.

ΔTf = 273.15 K - 271.15 K = 2 K

Now, let's calculate the freezing point depression for the 5% glucose solution.

ΔTf = Kf * m 2 K = Kf * 0.117 mol/kg

Finally, we can solve for the freezing point of the 5% glucose solution.

Freezing point of the 5% glucose solution = Freezing point of pure water - ΔTf = 273.15 K - 2 K = 271.15 K

Therefore, the freezing point of the 5% glucose solution in water is 271.15 K.