Chapter Two · Secondly: Quantitative analysis: practice questions
Chapter Two · Chemical analysis · Chemistry. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 84 questions on this lesson, each with an explanation.
Phenolphthalein indicator shows which colours in acidic and in alkaline media, respectively?
AColourless in acid and pink (red) in alkali.
BYellow in acid and colourless in alkali.
CRed in acid and yellow in alkali.
DBlue in acid and red in alkali.
Show answer and explanation
Correct answer: (A) Colourless in acid and pink (red) in alkali.
Explanation: Phenolphthalein is colourless in acidic solution and turns pink (red) in alkaline solution.
Question 2Easy
50 mL of 1.0 M sodium chloride solution is diluted with distilled water to a final volume of 250 mL. The molarity of the diluted solution is:
A5.0 M
B0.5 M
C0.1 M
D0.2 M
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Correct answer: (D) 0.2 M
Explanation: Using M₁V₁ = M₂V₂, M₂ = (1.0 × 50) / 250 = 0.2 M.
Question 3Easy
The molarity (M) of a solution is defined as the number of moles of solute dissolved in:
A100 mL of the solvent.
Bone kilogram of the solution.
Cone litre (1000 mL) of the solution.
D1000 grams of the solvent.
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Correct answer: (C) one litre (1000 mL) of the solution.
Explanation: Molarity is the number of moles of solute dissolved in one litre (1000 mL) of solution.
Question 4Medium
For the titration of a strong acid against a strong base, which of the following indicators can be used to detect the end point?
AEither methyl orange or phenolphthalein, because the pH changes sharply over a wide range at the equivalence point.
BOnly phenolphthalein, because the salt is alkaline.
COnly methyl orange, because the salt is acidic.
DOnly litmus, because it is a natural indicator.
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Correct answer: (A) Either methyl orange or phenolphthalein, because the pH changes sharply over a wide range at the equivalence point.
Explanation: In a strong acid–strong base titration the pH jump is large, so either methyl orange or phenolphthalein can be used.
Question 5Medium
What volume of distilled water must be added to 50 mL of 2 M hydrochloric acid to dilute it to 0.5 M? (assume volumes are additive)
A150 mL
B200 mL
C100 mL
D250 mL
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Correct answer: (A) 150 mL
Explanation: 2 M × 50 mL = 0.5 M × V2 → V2 = 200 mL total; water added = 200 − 50 = 150 mL.
Question 6Medium
A 0.585 g sample of pure sodium chloride is dissolved and treated with excess silver nitrate solution. The mass of silver chloride precipitate obtained is: (Na = 23, Cl = 35.5, Ag = 108)
A1.435 g
B0.585 g
C0.7175 g
D2.87 g
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Correct answer: (A) 1.435 g
Explanation: n(NaCl) = 0.585 / 58.5 = 0.01 mol = n(AgCl); mass = 0.01 × 143.5 = 1.435 g.
Question 7Medium
Which of the following indicators turns red in acidic medium and yellow in basic medium?
APhenolphthalein
BLitmus
CBromothymol blue
DMethyl orange
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Correct answer: (D) Methyl orange
Explanation: The indicator table shows that Methyl orange is Red in acidic medium and Yellow in basic medium.
Question 8Hard
An impure sample of sodium hydroxide of mass 0.50 g is dissolved in water and completely neutralised by 100 mL of 0.1 M hydrochloric acid (NaOH + HCl → NaCl + H2O; NaOH = 40 g/mol). What is the percentage purity of the sodium hydroxide?
When calculating the number of molecules of water of crystallization for anhydrous barium chloride combining with water, if the mass of anhydrous barium chloride is 2.2923 grams and the combined mass of water of crystallization is 0.398 grams, what is the exact percentage of water of crystallization?
A12.50 %
B14.79 %
C16.35 %
D18.20 %
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Correct answer: (B) 14.79 %
Explanation: % Water of crystallization = (0.398)/(2.6903) × 100 = 14.79 %.
Question 10Hard
In the derivation for hydrated barium chloride in the text, how many molecules of water of crystallization are found to combine with one formula unit of BaCl2?
A1 molecule
B3.5 molecules
C2.006 molecules
D4.1 molecules
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Correct answer: (C) 2.006 molecules
Explanation: 'The number of molecules of water of crystallization = (36.114)/(18) = 2.006 molecules', leading to BaCl2.2H2O.