← Back to topic
Vidaara.orgClass 11 · Chemistry
CodeVID-C11-01-T3-01
Concentration of Solutions — Practice Assignment
Chapter: Some Basic Concepts of Chemistry
Topic: Concentration of Solutions
Maximum Marks: 30
Time: 60 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Use atomic masses: H=1, C=12, O=16, Na=23, S=32, Cl=35.5.
  • Assume solution density = 1 g/mL unless stated otherwise.
  • Show all working and state units.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
The unit of molality is:
  • A.$mol\,L^{-1}$
  • B.$mol\,kg^{-1}$
  • C.$g\,L^{-1}$
  • D.dimensionless
2.
Molarity is defined per litre of:
  • A.solvent
  • B.solute
  • C.solution
  • D.water
3.
Which is used to express very dilute concentrations such as pollutants?
  • A.Molarity
  • B.Molality
  • C.ppm
  • D.Mole fraction
4.
The dilution equation is:
  • A.$M_1V_1 = M_2V_2$
  • B.$M_1V_2 = M_2V_1$
  • C.$M_1/V_1 = M_2/V_2$
  • D.$M_1 + V_1 = M_2 + V_2$
5.
Mass percent of solute equals:
  • A.(mass solute / mass solvent) x 100
  • B.(mass solute / mass solution) x 100
  • C.(moles solute / moles solution) x 100
  • D.(mass solvent / mass solution) x 100
Section B — Short Answer (2 marks) 3 × 2 = 6 marks
6.
Calculate the molarity of a solution with $0.2\,mol$ solute in $400\,mL$ of solution.
7.
Distinguish between molarity and molality in one line each.
8.
Find the mass percent of a solution containing $25\,g$ salt in $475\,g$ water.
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
9.
Calculate the molality of a solution prepared by dissolving $5.85\,g$ of $NaCl$ in $500\,g$ of water.
10.
Find the mole fraction of solute in a solution of $18\,g$ glucose (M=180) in $90\,g$ water.
Section D — Long Answer (5 marks) 1 × 5 = 5 marks
11.
A solution is prepared by dissolving $4\,g$ of $NaOH$ in water to make $200\,mL$. (a) Find its molarity. (b) What volume of this solution gives $0.01\,mol$ $NaOH$? (c) To what volume must $50\,mL$ of it be diluted to obtain $0.1\,M$?

Answer Key

Section A — Multiple Choice Questions
  1. (B) $mol\,kg^{-1}$
  2. (C) solution
  3. (C) ppm
  4. (A) $M_1V_1 = M_2V_2$
  5. (B) (mass solute / mass solution) x 100
Section B — Short Answer (2 marks)
  1. M = 0.2 / 0.4 = 0.5 mol/L.
  2. Molarity = moles of solute per litre of solution (temperature dependent). Molality = moles of solute per kilogram of solvent (temperature independent).
  3. Mass of solution = 500 g; % = (25/500) x 100 = 5%.
Section C — Short Answer (3 marks)
  1. M(NaCl) = 58.5; moles = 5.85/58.5 = 0.1 mol; solvent = 0.5 kg; molality = 0.1/0.5 = 0.2 m.
  2. Moles glucose = 18/180 = 0.1; moles water = 90/18 = 5; total = 5.1; x(glucose) = 0.1/5.1 = 0.0196.
Section D — Long Answer (5 marks)
  1. (a) moles NaOH = 4/40 = 0.1 mol; V = 0.2 L; M = 0.1/0.2 = 0.5 M. (b) 0.01 mol / 0.5 mol/L = 0.02 L = 20 mL. (c) M1V1 = M2V2: 0.5 x 50 = 0.1 x V2, V2 = 250 mL.
Generated by Vidaara.org · Assignment VID-C11-01-T3-01 · vidaara.org