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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 126p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 13 - Problem 126p

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# Answer: An aqueous solution containing 35.9 g of an unknown molecular (nonelectrolyte) ISBN: 9780321910295 34

## Solution for problem 126P Chapter 13

Introductory Chemistry | 5th Edition

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Problem 126P

Problem 126P

An aqueous solution containing 35.9 g of an unknown molecular (nonelectrolyte) compound in 150.0 g of water has a freezing point of −1.3 °C. Calculate the molar mass of the unknown compound.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the molar mass of the unknown compound.

Step1:

The freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.

In order to determine the molar mass of the solute, te following expression can be used:

Kf x w2 x 1000

M2 = ------------------------        -----(1)

ΔTf x w1

Where, M2 = molar mass of the solute

Kf = freezing point depression constant

w2 = given mass of solute in grams

ΔTf = depression in freezing point

w1 = mass of solvent in grams

Step2:

Freezing point of water = 0oC

Freezing point of the solution = -1.3oC

Therefore, depression in freezing point(ΔTf) = 0oC - (-1.3oC)

= 1.3oC

Mass of solvent(w1) = 150.0 g

Mass of solute(w2) = 35.9 g

Freezing point depression constant of water(Kf) = 1.86oC kg mol-1

Substituting the values in equation (1), we get,

1.86oC kg mol-1 x 35.9 g x 1000 g kg-1

M2 = ---------------------------------------------

1.3oC  x 150.0 g

= 66774 / 195 g mol-1

= 342.43 g mol-1

Thus, molar mass of the unknown compound is 342.43 g/mol.

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Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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