Free Solutions and Colloids MCQs with Answers

1. Osmotic pressure (7) does NOT depend on

• A. Molarity
• B. unoveal gas constant
• C. Temperature
• D. structure of solute particle

Explanation: osmotic pressure does not depend on structure of solute particles but it depend on molarity ,temperature,and solute concentration

2. The osmotic pressure of equimolar solutions of BaCl2, NaCl and sucrose will be in the order:

• A. Sucrose > NaCl > BaCl2
• B. Sucrose > BaCl2 > NaCl
• C. NaCl > BaCl2 > Sucrose
• D. BaCl2 > NaCl > Sucrose

Explanation: Osmotic pressure is a colligative property, and the more the number of particles more the osmotic pressure. π=iCRT Hence, BaCl2>NaCl>Sucrose since the concentration is the same and i for BaCl2 is 3, NaCl is 2 and sucrose is 1.

3. Consider NaCl produced in the equation: FeCl3+ NaOH → Fe(OH)3 + NaCl 175.5g of NaCl was dissolved in water to make a liter of solution, the molarity would be:

• A. 0.1 M
• B. 3 M
• C. 8 M
• D. 12 M
• E. 16 M

Explanation: Molarity = number of moles/volume in dm3 or liters 175.5 g make up 3 moles of NaCl 175.5/58.5 = 3 mol That means 3 mole of NaCl is dissolved in 1 liter of solution that makes the NaCl solution 3M.

4. An azeotropic mixture of two liquids has a boiling point higher than either of them when it:

• A. Shows positive deviation from Raoult's law
• B. Shows negative deviation from Raoult's law
• C. Shows ideal behaviour
• D. Is saturated

Explanation: An azeotropic solution is a solution that contains two liquids and has a constant vapour and liquid composition. The azeotropic solutions cannot be separated with the help of the distillation process. They are divided into two parts i.e. positive and negative deviation. Roult's law states that the vapour pressure of a solution that contains a non-volatile solute is directly proportional to the mole fraction of the solvent.

5. The number of moles of solute dissolved per liter of solution is called:

• A. Normality
• B. Molality
• C. Mole fraction
• D. Molarity
• E. Percentage composition

Explanation: Molarity (M) indicates the number of moles of solute per liter of solution.Molality is a property of a solution and is defined as the number of moles of solute per kilogram of solvent. Mole fraction is defined as the ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the moles in solution. Normality is described as the number of gram or mole equivalents of solute present in one liter of a solution. The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

6. From a mixture of CO2 and H2 gases, CO2 can be separated by passing the mixture through:

• A. Water at high temperature
• B. Water under high pressure
• C. Cold water
• D. Acidified water

Explanation: Carbon dioxide is separated from hydrogen by passing the mixture through the water under high pressure. When carbon dioxide is compressed into water at pressure, the CO2 gas dissolves into the liquid. When pressure is released, CO2 releases from the liquid as a gas in the form of gas bubbles. This process is also called carbonation and is the basis of many popular beverages. Since hydrogen will not dissolve in the water, the two gases will be effectively separated. Carbon dioxide will not dissolve in the forms of water given in all the remaining options, so separation can not occur, leaving only option B as the correct option.

7. The solubility product for BaSO4, at 18-25°C is:

• A. 1.0 x 10^-10 mol2dm-6
• B. 8.7 x 10^-36 mol2dm-6
• C. 1.8 x 10^-21 mol2dm-6
• D. 8.4 x 10^-28 mol2dm-6
• E. 3.5 x 10^-52 mol2dm-6

Explanation: The solubility product for BaSO4 represents the equilibrium constant for its dissolution reaction. At 18-25°C, BaSO4 has a very low solubility, resulting in a small concentration of Ba²+ and SO4²- ions in solution, which yields a solubility product of 1×10¹⁰mol²dm-6.

8. 0.5 molar solution NaOH contains:

• A. 40g NaOH in one dm3
• B. 80g NaOH in one dm3
• C. 10g NaOH in one dm3
• D. 20g NaOH in one dm3

Explanation: The atomic mass of Na is 23g/mol, the atomic mass of O is 16g/mol, and the atomic mass of H is 1g/mol so, by adding all three atomic masses, we can obtain the mass number of NaOH, which is 23+16+1=40g/mol. For a solution to be 1 molar, 1 mole of a substance has to be dissolved in 1 dm3 of the solvent. Since we have to find out the mass of NaOH needed to make a 0.5 molar solution, we will dissolve 0.5 moles of NaOH in 1 dm3 of the solvent. 0.5 mol of NaOH corresponds to 20g of NaOH which can be worked out using the proportion method that if 1 mol of NaOH contains 40g of NaOH (as calculated above) then 0.5 mol of NaOH will contain 20g of NaOH. If the proportion method seems difficult, the formula relating mass with moles can be used which is: n=m/M; where M is molar mass, n is the number of moles, and m is the mass of NaOH for n moles. (0.5) = m/(40) m=(40)(0.5)=20g.

9. The temperature at which one crystalline form changes to other is called _ temperature:

• A. Critical
• B. Absolute
• C. Transition
• D. None of these

Explanation: The temperature at which one crystalline form changes into another is called the transition temperature or phase transition temperature. This process is commonly observed in materials undergoing a phase transition, such as from a solid to another solid polymorphic form. The specific terms for these transitions can vary depending on the nature of the substance, for example, it could be called a melting point, freezing point, or transition temperature.

10. Aqueous solutions of Iodine and Sodium hydroxide were mixed in a round bottom flask at 70 C. Following chemical reaction was carried out. 3I2 + 6 NaOH → NaIO3 + NaI + H2OThis reaction is termed as:

• A. Redox reaction
• B. Precipitation reaction
• C. Substitution reaction
• D. Free radical reaction

Explanation: To determine whether such a reaction is a redox reaction, the oxidation states of the reactants and products is compared. The oxidation state of Na remains +1 throughout, as does the oxidation states of Hydrogen (+1) and Oxygen (-2). Iodine undergoes a change in oxidation state. Its oxidation state is 0, in its molecular form, on the reactant side. In NaI, iodine's oxidation state is -1 so, Iodine has undergone reduction. In NaIO3, the oxidation states of Na remain +1 and oxygen's also remains -2 so, we can determine that the oxidation state of iodine is +5 in this reaction. (NOTE: For oxidation numbers, the total oxidation state of a compound is zero so, Iodine's oxidation number must be +5 to ensure NaIO3's oxidation state is zero). Since Iodine underwent oxidation AND reduction, we can conclude that this is a redox reaction (disproportionation to be precise, as the same species has undergone oxidation and reduction) so, option A is correct. Option B is incorrect, as the two salts produced are soluble in water. Option C does not describe this reaction. Option D is incorrect as no free radicals are produced during this reaction.