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Test: Expressing Concentration of Solutions (NCERT) - NEET MCQ


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15 Questions MCQ Test Chemistry Class 12 - Test: Expressing Concentration of Solutions (NCERT)

Test: Expressing Concentration of Solutions (NCERT) for NEET 2024 is part of Chemistry Class 12 preparation. The Test: Expressing Concentration of Solutions (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Expressing Concentration of Solutions (NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Expressing Concentration of Solutions (NCERT) below.
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Test: Expressing Concentration of Solutions (NCERT) - Question 1

What is the mass percentage of carbon tetrachloride if 22g of benzene is dissolved in 122g of carbon tetrachloride?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 1

Mass of solution = Mass of C6H6 + Mass of CCl4
= 22 + 122 = 144 g
Mass % of CCl4= 84.72%

Test: Expressing Concentration of Solutions (NCERT) - Question 2

What is the mole fraction of glucose in 10% w/W glucose solution?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 2

No. of moles of glucose = 10/180 = 0.0555 mol
No. of moles of water = 90/18 = 5 mol
Number of moles of solution = 5.0555 mol
Mole fraction of glucose = No. of moles of glucose/No. of moles of solution = 0.0555/5.0555 = 0.01

Test: Expressing Concentration of Solutions (NCERT) - Question 3

Calculate the percentage composition of a solution obtained by mixing 300 g of a 20% and 200 g of a 30% solution by weight.

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 3

Mass of solute is obtained by multiplying % weight with mass of solvent and obtaining an answer divided by 100.
Mass of solute in 200g of 30% solution can be obtained as follows:
30% by weight indicates 30g of solute is present in 100g of solvent so 60g of solute will be present in 200g of solution.
20% by weight indicates 20g of solute is present in 100g of solvent so 60g of solute will be present in 300g of solution
So on mixing these both solutions, mass of solute is 60g +60g =120g
Mass of solution on mixing is 200g +300g =500g
Percent of solute can be obtained as follows:

24% percentage of solute in the final solution.

Test: Expressing Concentration of Solutions (NCERT) - Question 4

When 1.04g of BaCl2 is present in 105g of solution, the concentration of solution is:

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 4

ppm = 

= 10.4 ppm

Test: Expressing Concentration of Solutions (NCERT) - Question 5

What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 5


Mass of C2H5OH = 85 g
Molar mass of C2H5OH = 46 g/mol
nC2H5OH = 85/46 = 1.85 mol
Mass of water = 100 - 85 = 15 g
nH2O = 15/18 = 0.833 mol

Test: Expressing Concentration of Solutions (NCERT) - Question 6

What is the molarity of a solution containing 10 g of NaOH in 500 mL of solution?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 6

No. of moles of NaOH = 10/40 = 0.25 mol

Test: Expressing Concentration of Solutions (NCERT) - Question 7

What will be the molarity of 30mL of 0.5M H2SO4 ​solution diluted to 500mL?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 7

V1 = 30 mL, M= 0.5 M, V2 = 500 mL, M2 = ?,
M1V1 = M2V2
0.5 x 30 = M2 × 500 or M2 = 0.03 M

Test: Expressing Concentration of Solutions (NCERT) - Question 8

How many Na+ ions are present in 100 mL of 0.25 M of NaCl solution?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 8


NaCl → Na+ Cl
No. of moles of Na+ ions = 0.025
No. of Na+ ions = 0.025 x 6.023 x 1023 = 1.505 × 1022

Test: Expressing Concentration of Solutions (NCERT) - Question 9

How many grams of NaOH are present in 250 mL of 0.5 M NaOH solution?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 9

No. of moles of NaOH

Mass of NaOH = 40 × 0.125 = 5g

Test: Expressing Concentration of Solutions (NCERT) - Question 10

250mL  of sodium carbonate solution contains 2.65g of Na2CO3. If 10mL of this solution is diluted to 500mL, the concentration of the diluted acid will be

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 10

Molarity of Na2CO3 solution
= 2.65/106 × 1000/250 = 0.1 M
10mL of this solution is diluted to 500mL
M1V1 = M2V2 
0.1 × 10 = M2 × 500
⇒ M2 = 0.002M

Test: Expressing Concentration of Solutions (NCERT) - Question 11

The density of a solution prepared by dissolving 120g of urea (mol. mass = 60u) in 1000g of water is 1.15g/mL. The molarity of this solution is

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 11

Mass of solute taken = 120g
Molecular mass of solute = 60u
Mass of solvent = 1000g
Density of solution = 1.15g/mL
Total mass of solution = 1000 + 120 = 1120g

Test: Expressing Concentration of Solutions (NCERT) - Question 12

What will be the molality of a solution of glucose in water which is 10% w/W?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 12

Mass of the solution = 100g
Mass of glucose = 10g, Mass of water = 90 g
No. of moles of glucose = 10/180 = 0.0555mol
No. of moles of water = 90/18 = 5 mol
Molality = No. of moles of solute/Mass of solvent in kg
= 0.0555mol/0.0090kg = 0.617m

Test: Expressing Concentration of Solutions (NCERT) - Question 13

The molality of 648 g of pure water is

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 13

Molality = no of moles of solute/mass of solvent in kg
Molar mass of water = 18 g
Mass of water = 648 g
No of moles of water = 648/18 = 36 mol

Test: Expressing Concentration of Solutions (NCERT) - Question 14

What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution?

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 14

Mass of solvent = 1000 g
Molar mass of urea (NH2CONH2) = 60gmol−1
0.25 mole of urea = 0.25 × 60 = 15g
Total mass of solution = 100 + 15 = 1.015kg
1.015 kg of solution contain urea = 15g
2.5 kg of solution = 

Test: Expressing Concentration of Solutions (NCERT) - Question 15

Concentration terms like mass percentage, ppm, mole fraction and molality do not depend on temperature. However, molarity is a function of temperature because

Detailed Solution for Test: Expressing Concentration of Solutions (NCERT) - Question 15

Molarity of a solution is defined as the number of moles of solute dissolved per litre of solution. Since volume depends on temperature and changes with change in temperature, therefore, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, and therefore, concentration terms such as mass percentage, mole fraction and molality which do not involve volume are independent of temperature.

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