Free Solutions and Colloids MCQs with Answers

111. What is the ionic strength of 0.01 M Barium Chloride solution?

• A. 0.03
• B. 0.02
• C. 0.04
• D. 0.01

Explanation: We know that:Ionic strength (I) = ½ (m1z1² + m2z2²)Where,m1 = concentration of Ba2+ = 0.01m2 = concentration of Cl- = 0.01 × 2 = 0.02z1 = Charged on Ba2+ = 2z2 = Charged on Cl- = -1Putting the values in the formula of ionic strength, we get:I = ½ (0.01 × 2² + 0.02 × (-1)²)= ½ (0.01 × 4 + 0.02 × 1)= ½ (0.04 + 0.02)= ½ (0.06)I = 0.03

112. 2.3g of ethanol (C2H5OH) is added to 500g of water to determine the molality of the resulting solution;

• A. 0.01 molal
• B. 0.1 molal
• C. 1.1 molal
• D. 1.0 molal

Explanation: Mass of ethanol (solute) = 2.3gMass of water (solvent) = 500gFirst, we need to convert the mass of ethanol to moles. The molar mass of ethanol is approximately 46.07 g/molNumber of moles of ethanol = Mas of ethanol / molar mass of ethanolNumber of moles of ethanol = 2.3 / 46.07 = 0.05 molWe need to convert the mass of water to kilograms. So,Mas of water = 500 g = 500 / 1000 = 0.5 kgFinally, we can calculate the molality of the solution:Molality = Number of moles of ethanol/mass of water in kilogramsMolality = 0.05 mol / 0.5kg = 0.1 mol/kgTherefore, the molality of the resulting solution is 0.1 molal.

113. If we pass current through the sucrose solution the galvanometer will not show any deflection because sucrose molecules:

• A. Move towards cathode
• B. Move towards anode
• C. React with water
• D. Remain neutral

Explanation: Sucrose molecules are electrically neutral, meaning they have an equal number of positive and negative charges. Therefore, when current is passed through a sucrose solution, the molecules will not move towards the cathode or anode, nor will they react with water. As a result, the Galvanometer will not show any deflection. However, if the sucrose solution contains ions or other charged particles, the Galvanometer will show deflection as these charged particles will move toward the oppositely charged electrode.

114. 50 cm of KOH solution was titrated against 1.0M HCIusing phenolphthalein as an indicator. The acid used was found to be 7,5 cm. the concentration of KOH solution is:

• A. 0.15 M
• B. 1.5 M
• C. 0.75
• D. None

Explanation: The calculations are shown below : The chemical equation used in this titration is : HCl+KOH→KCl+H₂O In this chemical reaction, the molar ratio between HCl and KOH is 1:1. So, moles of HCl = moles of KOH Mʜᴄʟ × Vʜᴄʟ = Mᴋᴏʜ × Vᴋᴏʜ Mʜᴄʟ × Vʜᴄʟ ----------------- = Mᴋᴏʜ Vᴋᴏʜ (placing values) 1 × 7.5 ------------ = Mᴋᴏʜ 50 Mᴋᴏʜ = 0.15M

115. The sum of mole fractions of solute and solvent is always equal to:

• A. 0.1
• B. 10.0
• C. 1.0
• D. Zero

Explanation: The fundamental concept of mole fractions dictates that the sum of the mole fractions of all components in a solution, including both solute and solvent, must equal 1. This is because mole fraction is a way of expressing the concentration of a component in a mixture, and all component concentrations together must sum to the whole, which is 1.0. Options A, B, and D are incorrect because they do not satisfy this fundamental principle: 0.1 and zero are too low, and 10.0 is too high.

116. If 20.0 cm3 of 0.5 M solution is diluted to 1.0 dm3. What will be its new concentration?

• A. 0.001 M
• B. 0.01 M
• C. 1.0 M
• D. 10.0 M

Explanation: To find the new concentration after dilution, use the formula M1V1 = M2V2. Here, M1 is the initial concentration (0.5 M), V1 is the initial volume (20.0 cm3 or 0.02 dm3), and V2 is the final volume (1.0 dm3). Solving for M2 gives us M2 = (0.5 M * 0.02 dm3) / 1.0 dm3, resulting in a concentration of 0.01 M. The other options are incorrect because they either assume incorrect calculations or misunderstand the concept of dilution.

117. Aiman in laboratory dissolve 4g of NaOH in 250ml of water. The molarity of this solution is:

• A. 0.4M
• B. 4M
• C. 0.2M
• D. 0.1M

Explanation: M = 4/40 / 250/1000 = 0.4,where 4g is the amount of NaOH, 40 is the grams per mole of NaOH, 250mL is the amount of water and 1000 is the number of ml in a liter.The most common way to express solution concentration is molarity (M), which is defined as the amount of solute in moles divided by the volume of solution in liters: M = moles of solute/liters of solution.

118. A water sample contains 3.8 x 10^3g of mercury per kilo gram of the sample. What is the concentration of mercury in parts per million?

• A. 3.8 ppm
• B. 38 ppm
• C. 0.38 ppm
• D. 380 ppm

Explanation: The concentration of mercury is calculated by the formula:ppm = (mass of solute / mass of solution) x 106Given that you have 3.8 x 103g of mercury per kilogram of the sample, first convert this to the same units: 3.8 x 103 g/kg is equivalent to 3.8 g/kg. Since 1 kg equals 1,000,000 mg, the concentration is (3.8 / 1000) x 106 = 3.8 ppm.Other options are incorrect due to calculation errors or misinterpretation of the conversion factor to ppm.

119. 10.0 dm3 gas cylinder containing mixture of various gases 50cm3 of nitrogen gas is in the mixture what is the concentration of N2 gas in part per billion (ppb).

• A. Option A
• B. Option B
• C. Option C
• D. Option D

Explanation: Students should remember that a billion has 9 zeros so option C and D can be eliminated. Also that 1 dm³ = 1000 cm³ so 10 dm³ = 10,000 cm³ So the correct answer is option B i.e 50/10,000 x 109As it is numerical, it can have only one answer.