A0.1539 molar aqueous solution of cane sugar (mol. mass =342gmol -1)has an . f pt .of 271K.What will be . f pt . Of an aqueous solution containing 5g of glucose(molar mass =180g mol -1) per 100g of solution? ( . f pt .of H 2O=273.15K)

To determine the freezing point of the aqueous solution containing glucose, we can use the concept of freezing point depression. This phenomenon occurs when a solute is added to a solvent, resulting in a lower freezing point than that of the pure solvent. The formula for freezing point depression is given by:

Freezing Point Depression Formula

The formula is:

ΔTf = i * Kf * m

• ΔTf = change in freezing point
• i = van 't Hoff factor (for glucose, i = 1 since it does not dissociate)
• Kf = freezing point depression constant of the solvent (for water, Kf = 1.86 °C kg/mol)
• m = molality of the solution (moles of solute per kg of solvent)

Step 1: Calculate the Molality of the Glucose Solution

First, we need to find the number of moles of glucose in the solution. The molar mass of glucose is 180 g/mol. Given that we have 5 g of glucose, we can calculate the moles:

moles of glucose = mass (g) / molar mass (g/mol)

moles of glucose = 5 g / 180 g/mol = 0.02778 mol

Step 2: Determine the Mass of the Solvent

Next, we need to find the mass of the solvent (water) in kilograms. Since the total mass of the solution is 100 g and we have 5 g of glucose, the mass of water is:

mass of water = total mass of solution - mass of glucose

mass of water = 100 g - 5 g = 95 g = 0.095 kg

Step 3: Calculate the Molality

Now we can calculate the molality (m) of the solution:

molality (m) = moles of solute / mass of solvent (kg)

molality (m) = 0.02778 mol / 0.095 kg = 0.2925 mol/kg

Step 4: Calculate the Freezing Point Depression

Now we can use the freezing point depression formula:

ΔTf = i * Kf * m

Substituting the values:

ΔTf = 1 * 1.86 °C kg/mol * 0.2925 mol/kg = 0.54435 °C

Step 5: Determine the New Freezing Point

Finally, we need to find the new freezing point of the solution. The freezing point of pure water is 273.15 K, and we subtract the change in freezing point:

New freezing point = freezing point of pure water - ΔTf

New freezing point = 273.15 K - 0.54435 °C

Since we need to convert °C to K, we can directly subtract:

New freezing point = 273.15 K - 0.54435 K = 272.60565 K

Final Result

The freezing point of the aqueous solution containing 5 g of glucose per 100 g of solution is approximately 272.61 K.