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GPAT Solutions — practice questions — page 2

Explore 46 free Solutions questions in GPAT Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Solutions questions and exercises with answers

Showing questions 31–46 of 46.

A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:Read the answer and worked solution →Mole fraction of solute in 18% glucose (180 g/mol) by mass in water:Read the answer and worked solution →Vapour pressure of pure water is 100 (arbitrary units). Add 1 mole non-volatile solute to 9 moles water. New vapour pressure is:Read the answer and worked solution →Fill in the molality calculationRead the answer and worked solution →Complete the two concentration definitionsRead the answer and worked solution →Molarity or molality?Read the answer and worked solution →True or false: concentrationRead the answer and worked solution →Naming the measureRead the answer and worked solution →Complete the meaning of colligativeRead the answer and worked solution →True or false: colligative propertiesRead the answer and worked solution →Tap every statement true of molalityRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Flip and pair: measure with its denominatorRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →

Worked practice questions

A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:

  1. 1.5 m
  2. 1 m (approx) using ΔT/K_b
  3. 2 m
  4. 0.5 m
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Correct answer: B. 1 m (approx) using ΔT/K_b

We have ΔT_b = K_b × m → 0.5. So ΔT_b = 0.52 × m → m ≈ 0.96 ≈ 1 m. (Assuming non-electrolyte, i = 1.)

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Mole fraction of solute in 18% glucose (180 g/mol) by mass in water:

  1. 0.02
  2. 0.03
  3. 0.1
  4. 0.05
Show answer and solution

Correct answer: A. 0.02

**In 100 g solution:** 18 g glucose (= 0.1 mol), 82 g water (= 82/18 = 4.56 mol). x_glucose = 0.1/(0.1 + 4.56) = 0.1/4.66 ≈ 0.0215 ≈ 0.02.

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Vapour pressure of pure water is 100 (arbitrary units). Add 1 mole non-volatile solute to 9 moles water. New vapour pressure is:

  1. 91
  2. 10
  3. 100
  4. 90
Show answer and solution

Correct answer: D. 90

x_water = 9/10 = 0.9. By Raoult's law: P = 0.9 × 100 = 90 units. Lowering by 10 units (vapour pressure lowering).

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