(a) Suppose we dissolve a 6.00-g sample of a mixture of naphthalene, C10 H8, and anthracene, C14

step 1 Good day. The topic is about collicative properties. When solute is added to solvent, the proper

(a) Suppose we dissolve a 6.00-g sample of a mixture of...

(a) Suppose we dissolve a $6.00-\mathrm{g}$ sample of a mixture of naphthalene, $\mathrm{C}_{10} \mathrm{H}_{8},$ and anthracene, $\mathrm{C}_{14} \mathrm{H}_{10},$ in $360 . \mathrm{g}$ of benzene. The solution is observed to freeze at $4.85^{\circ} \mathrm{C}$. Find the percent composition (by mass) of the sample. (b) At what temperature should the solution boil? Assume that naphthalene and anthracene are nonvolatile nonelectrolytes.

(a) Suppose we dissolve a $6.00-\mathrm{g}$ sample of a mixture of naphthalene, $\mathrm{C}_{10} \mathrm{H}_{8},$ and anthracene, $\mathrm{C}_{14} \mathrm{H}_{10},$ in $360 . \mathrm{g}$ of benzene. The solution is observed to freeze at $4.85^{\circ} \mathrm{C}$. Find the percent composition (by mass) of the sample.
(b) At what temperature should the solution boil? Assume that naphthalene and anthracene are nonvolatile nonelectrolytes.

Key Concepts

Mixture Composition Analysis

This concept involves determining the percentage composition (by mass or moles) of components in a mixture using experimental data and properties like the changes in freezing or boiling points. Application of colligative property equations allows for the deconvolution of contributions from different components in the mixture to the overall observed physical property changes.

Colligative Properties

These are properties of solutions that depend solely on the number of solute particles relative to solvent particles, not on their identity. Colligative properties include phenomena such as freezing point depression and boiling point elevation, which are directly related to the solute’s concentration in the solvent.

Freezing Point Depression

This is a colligative property where the addition of a solute to a solvent lowers its freezing point. It is quantitatively described by the equation ?Tf = Kf × m, where ?Tf is the decrease in freezing temperature, Kf is the cryoscopic constant of the solvent, and m is the molality of the solution.

Boiling Point Elevation

Another colligative property, boiling point elevation refers to the increase in the boiling point of a solvent when a nonvolatile solute is dissolved in it. The effect is calculated by the relation ?Tb = Kb × m, where ?Tb is the elevation in boiling temperature, Kb is the ebullioscopic constant, and m is the molality.

Molality

Molality is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It is especially useful in colligative property calculations because it is temperature-independent, making it a reliable unit for quantifying solute concentration in a solution.

Transcript

00:01 Good day.

00:01 The topic is about collicative properties.

00:04 When solute is added to solvent, the properties of the solution that is formed are different from the properties of the pure solvent.

00:11 In particular, the solution freezes at a lower temperature and it boils at a higher temperature.

00:17 When a non -electrolitic and non -volatile solute is added to solvent, the solution has a freezing point and boiling point that is solved as follows.

00:27 We note that kf is the freezing point constant and kb, is the boiling point constant, which values depend on the type of solvent that is used.

00:35 And m is the molality of the solution.

00:38 Now let us suppose a non -volatile, non -electrolite naphthalene and anthracine, which total mass is 6 gram, is dissolved in 360 grams of benzene.

00:51 Benzine serves as our solvent.

00:57 So we let x as the mass of naphthalene.

01:03 And y as the mass of, so naftalin is n, and y is the mass of anthracine.

01:16 And it is given that the sum, or the total mass of these two, is six grams.

01:24 It is dissolved in 360 grams of benzene, or in kilograms, that's 0 .360 .0.

01:33 Now it was found out that as a result the solution freezes at 4 .85 degrees celsius.

01:40 So we say the f of the solution is 4 .85 degrees celsius.

01:48 So what we wish to find here is the percent composition, or that is the percent of naphthalene, and percent of anthracine in the original mixture.

01:59 Also, we also wish to find that boiling point of the solution.

02:06 So let us start by solving the molality of the solution.

02:13 So again, we have tf solution equals tf of solvent, which in this case is benzene minus kf times m, where the tf of the solution is 4 .85 degrees celsius.

02:29 Benzene freezes at a temperature of 5 .5 degrees celsius and it has a kf of 5 .12 degrees celsius per molal so that we will be able to solve for m here as follows.

02:49 4 .85 minus 5 .5 all over negative 5 .122 or this is equal to 0 .127 molal.

03:03 Hence, the molality of the solution is equal to .127 molal.

03:08 Now, let us find, or let us define that molality is equal to the most of solute over the kilograms of solvent, which in this case is benzene.

03:18 So we will be able to solve for the most absolute by multiplying the molality by the kilogram of benzene.

03:29 So that is 0 .127 times 0 .360.

03:37 And that is equal to 0 .057.

03:45 Note that this most of solute is the combination of the most of lafdalene and antacine.

03:52 That is most of n plus most of a...

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