Solubility and Intermolecular Forces

Solubility and Intermolecular Forces

Assessment

Interactive Video

Chemistry, Science, Physics

9th - 12th Grade

Practice Problem

Hard

Created by

Patricia Brown

FREE Resource

The video tutorial by Mr. Donnie covers factors affecting solubility through practice problems. It explains why NaCl is not soluble in benzene due to ionic and non-polar interactions, and identifies substances soluble in CCl4 and water based on polarity and hydrogen bonding. The effect of pressure on gas solubility is discussed, along with calculating Henry's Law constant for CO2. The tutorial also covers determining saturation levels of solutions and analyzing potassium chlorate solubility using a solubility chart.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why does NaCl have low solubility in benzene?

Both NaCl and benzene are ionic.

NaCl is ionic and benzene is non-polar.

NaCl is non-polar and benzene is polar.

Both NaCl and benzene are polar.

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which of the following is most soluble in CCl4?

NaCl

H2O

C10H22

NH3

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What type of intermolecular forces does CCl4 primarily exhibit?

Ionic bonds

Hydrogen bonding

Dipole-dipole interactions

London dispersion forces

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which substance is most likely to dissolve in water?

CH2CH2OH

CH3(CH2)8CH2OH

CHCl3

CCl4

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What characteristic of water makes it a good solvent for polar substances?

Its high density

Its ionic nature

Its ability to form hydrogen bonds

Its non-polar nature

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does pressure primarily affect in terms of solubility?

Solubility of solids

Solubility of liquids

Solubility of gases

Solubility of ionic compounds

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is Henry's law constant calculated?

By multiplying the concentration of the gas by the partial pressure

By dividing the partial pressure by the concentration of the gas

By dividing the concentration of the gas by the partial pressure

By adding the concentration of the gas to the partial pressure

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