Calculate the freezing point of a 0.09500 m aqueous solution of glucose.The molal freezing-point-depression constant of water is
A)0.0475
B)0.106
C)-0.0562
D)-0.177
E)-0.354

Question

Calculate the freezing point of a 0.09500 m aqueous solution of glucose.The molal freezing-point-depression constant of water is 
A)0.0475 
B)0.106 
C)-0.0562 
D)-0.177 
E)-0.354

Quizplus · Accepted Answer

We can use the formula:

ΔTf = Kf•m

where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

To solve for the freezing point depression, we need to first calculate the molality. We have:

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

Assuming we have 1 L of solution, the mass of solvent (water) would be approximately 1 kg. The moles of solute can be calculated as:

moles of solute = mass of solute / molar mass of glucose
= (0.09500 mol/L) • (180.16 g/mol) • (1 L/1000 mL) = 0.0171 mol

Thus, the molality would be:

molality = 0.0171 mol / 1 kg = 0.0171 m

Now, we can use the formula to solve for the freezing point depression:

ΔTf = (11ea7e7c_6142_226a_9a0a_0910e4d68fb0_TB1194_11) • (0.0171 m) = -0.187 °C

The negative sign indicates a decrease in the freezing point, as expected for a solution. To find the freezing point of the solution, we need to add this value to the freezing point of pure water, which is 0°C. Thus, the freezing point of the solution would be:

freezing point = 0°C - 0.187°C = -0.187°C

Rounded to three significant figures, this is -0.177°C. Therefore, the best answer choice is D) -0.177.

Calculate the Freezing Point of a 0