A student determined the molar mass of an unknown...

A student determined the molar mass of an unknown non-dissociating liquid by the method described in this experiment. She found that the equilibrium temperature of a mixture of ice and pure water was indicated to be - 0.1 $^{\circ}$ C on her thermometer. When she added $9.9 \mathrm{g}$ of her sample to the mixture. the temperature, after thorough stirring, fell to $-3.7^{\circ} \mathrm{C}$. She then poured off the solution through a screen into a beaker. The mass of the solution was $84.2 \mathrm{g}$. a. What was the freezing point depression? ______ $\mathrm{C}^{\circ}$ b. What was the molality of the unknown liquid? ______ $\mathbf{m}$ c. What mass of unknown liquid was in the decanted solution? d. What mass of water was in the decanted solution? ______ $\mathbf{g}$ e. How much unknown liquid would there be in a solution containing $1 \mathrm{kg}$ of water, with the student's unknown liquid at the same concentration as she had in her experiment? ______ g unknown liquid f. Based on these data, what value did she calculate for the molar mass of her unknown liquid, assuming she carried out the calculation correctly? ______ $g / m o l$

A student determined the molar mass of an unknown non-dissociating liquid by the method described in this experiment. She found that the equilibrium temperature of a mixture of ice and pure water was indicated to be - 0.1 $^{\circ}$ C on her thermometer. When she added $9.9 \mathrm{g}$ of her sample to the mixture. the temperature, after thorough stirring, fell to $-3.7^{\circ} \mathrm{C}$. She then poured off the solution through a screen into a beaker. The mass of the solution was $84.2 \mathrm{g}$.
a. What was the freezing point depression?
______ $\mathrm{C}^{\circ}$
b. What was the molality of the unknown liquid?
______ $\mathbf{m}$
c. What mass of unknown liquid was in the decanted solution?
d. What mass of water was in the decanted solution?
______ $\mathbf{g}$
e. How much unknown liquid would there be in a solution containing $1 \mathrm{kg}$ of water, with the student's unknown liquid at the same concentration as she had in her experiment?
______ g unknown liquid
f. Based on these data, what value did she calculate for the molar mass of her unknown liquid, assuming she carried out the calculation correctly?
______ $g / m o l$

Key Concepts

Freezing Point Depression

Freezing point depression is the decrease in the freezing temperature of a solvent when a non-volatile solute is added. This effect is quantified by the equation ?Tf = Kf × m, where ?Tf is the freezing point depression, Kf is the molal freezing point depression constant of the solvent, and m is the molality of the solute. This concept is widely used in experiments to determine solute concentrations and, indirectly, their molar masses.

Molality

Molality is a concentration unit defined as the number of moles of solute per kilogram of solvent. Unlike molarity, it is independent of temperature because it is based on mass rather than volume, making it particularly useful in experiments involving temperature changes such as freezing point depression studies.

Colligative Properties

Colligative properties are characteristics of a solution that depend on the number of solute particles present rather than the chemical identity of those particles. They include phenomena such as freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure, and are useful for determining various properties of solutes in a solvent.

Molar Mass Determination via Colligative Properties

Molar mass determination using colligative properties involves using measurements like freezing point depression to calculate the molality of a solution. By measuring the change in freezing point and knowing the Kf value of the solvent, one can compute the moles of solute per kilogram of solvent. Combined with the mass of solute added, this information allows for the calculation of the solute’s molar mass.

Transcript

00:02 Okay, here we have a freezing point depression problem.

00:05 And we're told that we have a non -dissociating liquid, which means our vantaf factor will be one.

00:23 So i know i equals one.

00:27 And we're found that ice water has a freezing point of negative 0 .1 degrees c.

00:40 And the liquid, the unknown liquid, and the ice water has a freezing point of negative 3 .7 degrees c.

01:01 Okay, so our first answer that we're going to get here before.

01:11 There we go.

01:13 My delta t is going to equal negative 3 .7 degrees c minus negative 0 .1 degrees c, and that will equal 3 .6.

01:41 What is the freezing point? degrees c, is my freezing point depression? and then b, we're asked for a molality, and to find the molality, we'll use delta t equals our freezing constant times molality times i, or i equals one.

02:13 So formalality will equal my delta t divided by kf, so molality will equal 3 .6 degrees c, divided by 1 .86.

02:30 Make sure this is right here.

02:39 Excuse me.

02:53 And that'll be degrees c over morality.

02:58 And that'll be equal to without a morality will equal grabbing my calculator.

03:06 Just sort of unwieldy because of charging.

03:10 3 .6 divided by 1 .86 is 1 .94.

03:25 I'll run later.

03:26 1 .935.

03:31 So our molality would be equal to 1 .9 to the correct number of sikvakes.

03:40 C asks us to find what mass of unknown liquid is in the dissolved sample.

04:02 And that will be what we were given.