Question

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated molar mass of the solute have been too high or too low? Explain.

Select all that apply.

Select all that apply.

The resulting molar mass (g/mol) would be smaller since the mol term is in the denominator.

A larger ΔT results in a lower calculated molality of the solution, and thus smaller number of moles.

The resulting molar mass (g/mol) would be larger since the mol term is in the denominator.

A larger ΔT results in a higher calculated molality of the solution, and thus larger number of moles.

If the temperature were 0.3 oC lower than the true value, the ΔT would be larger.

If the temperature were 0.3 oC lower than the true value, the ΔT would be smaller.

Answer 1

The formula for depression in freezing point is : -

where is the depression in freezing point

i.e. = Freezing point of Solvent - Freezing point of Solution(solvent + solute)

is the molal freezing point depression constant

and is the molality

Now, If the freezing point of the solution had been incorrectly read 0.3° C lower than the true freezing point, i.e. Freezing point of Solution(solvent + solute) is incorrectly read 0.3° C lower

(Freezing point of Solvent - Freezing point of Solution) becomes larger than true value of

is in the denominator and is larger than it's true value (as proved above)

Calculated molar mass of solute will be smaller i.e. too low

Now coming to the statements in the question : -

(1)

Seeing the formula above : -

i.e. they are inversely proportional to each other

The resulting molar mass (g/mol) would be smaller, if  is higher & vice-versa.

Therefore, 1st statement applies successfully

(2)

Therefore, A larger ΔT results in a higher calculated molality of the solution

and, because molality is directly proportional to Moles of solute, thus resulting in larger number of moles.

Therefore 2nd statement is false.

(3)

This statement is opposite of 1st statement, therefore it is false.

(4)

This statement is proved in (2) point (where 2nd statement came out to be false).

Therefore 4th statement applies successfully.

(5)

  =  Freezing point of Solvent - Freezing point of Solution(solvent + solute)

If the temperature were 0.3° C lower than the true value, i.e. Freezing point of Solution(solvent + solute) is 0.3° C lower than the true value, The ΔT would be larger.

Hence 5th statement applies successfully and 6th being exactly opposite is false.