   Chapter 11, Problem 73E

Chapter
Section
Textbook Problem

# a. Calculate the freezing-point depression and osmotic pressure at 25°C of an aqueous solution containing 1.0 g/L of a protein (molar mass = 9.0 × 104 g/mol) if the density of the solution is 1.0 g/cm3.b. Considering your answer to part a, which colligative property, freezing-point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

(a)

Interpretation Introduction

Interpretation:

The freezing-point depression, osmotic pressure and molar mass of given solution has to be explained.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution. These properties have a direct relationship to the solute particles, and therefore the colligative properties are useful for identifying the nature of solute particles and also calculating the molar masses of substances.

The flow of solvent through a semi-permeable membrane into the solution is called as osmosis. By the time system reaches equilibrium, the changes in the liquid level stops. There is a higher hydrostatic pressure on the solution than compared to that of the pure solvent because there is variation in the liquid levels at this point. The excess pressure on the solution is called osmotic pressure.

The osmotic pressure of solution is calculated by using,

π=MRT

Here,

π=osmoticpressure(in atm)

M=molarityofsolution(inM)

R=GasLawconstant Latm

T=Temperature(inK)

The change in depression in the freezing point expressed by the given equation

ΔTf=Kfm

Here Kf is molal freezing-point depression constant.

ΔTf is freezing-point depression, m is a molality

Explanation

Record the given info

Molar mass of protein= 9.0×104g/mol

Weight of protein= 1.0g

The Molar mass of protein and Weight of protein of aqueous solution is recorded as shown above.

To calculate the molarity of solution

M=1.0gproteinL×1mol9.0×104g=1.1×10-5mol/L

Molarity of solution

= 1.1×10-5mol/L

The molarity of solution is calculated by plugging in the values of molar mass, weight of protein.  The molarity of solution is found to be =1.1×10-5mol/L

To determine: The osmotic pressure of given solution

To calculate the osmotic pressure of given solution

Molarity= 1.1×10-5mol/L

Temperature=298K

π=1.1×10-5molL×0.08206LatmK mol=298K×760torratmπ=0

(b)

Interpretation Introduction

Concept Introduction:

Molar mass is expressed by

molar mass=massofsubstanceamountofsubstance

To Determining the molar mass of large molecules.

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