Solutions Test

Result

Solutions Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

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

A
84.72%

B
15.28%

C
50%

D
44%

Solution

Mass of solution = Mass of C₆H₆ + Mass of CCl₄
= 22 + 122 = 144 g
Mass % of CCl₄ = = 84.72%
Question 2
4 / -1

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

A
0.01

B
0.02

C
0.03

D
0.04

Solution

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
Question 3
4 / -1

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

A
50%

B
28%

C
64%

D
24%

Solution

Solute in 200 g of 30% solution = 60g
Solute in 300 g of 20% solution = 60 g
Total grams of solute = 120 g
Total grams of solution = 200 + 300 = 500g
% of solute in the final solution =
Question 4
4 / -1

When 1.04g of BaCl₂is present in 10⁵g of solution, the concentration of solution is:

A
0.104ppm

B
10.4ppm

C
0.0104ppm

D
104ppm

Solution

ppm =

= 10.4 ppm
Question 5
4 / -1

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

A
0.69

B
0.82

C
0.85

D
0.60

Solution

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
Question 6
4 / -1

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

A
0.25 mol L⁻¹

B
0.75 mol L⁻¹

C
0.5 mol L⁻¹

D
1.25 mol L⁻¹

Solution

No. of moles of NaOH = 10/40 = 0.25 mol
Question 7
4 / -1

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

A
0.3 M

B
0.03 M

C
3 M

D
0.103 M

Solution

V₁ = 30 mL, M₁= 0.5 M, V₂ = 500 mL, M₂ = ?,
M₁V₁ = M₂V₂
0.5 x 30 = M₂ × 500 or M₂ = 0.03 M
Question 8
4 / -1

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

A
0.025 × 10²³

B
1.505 × 10²²

C
15 × 10²²

D
2.5 × 10²³

Solution

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
Question 9
4 / -1

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

A
7.32 g

B
3.8 g

C
5g

D
0.5 g

Solution

No. of moles of NaOH

Mass of NaOH = 40 × 0.125 = 5g
Question 10
4 / -1

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

A
0.01 M

B
0.001 M

C
0.05 M

D
0.002 M

Solution

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
Question 11
4 / -1

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

A
1.78M

B
1.02M

C
2.05M

D
0.50M

Solution

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
Question 12
4 / -1

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

A
0.01 m

B
0.617 m

C
0.668 m

D
1.623 m

Solution

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
Question 13
4 / -1

The molality of 648 g of pure water is

A
36 m

B
55.5 m

C
3.6 m

D
5.55 m

Solution

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
Question 14
4 / -1

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

A
37 g

B
25 g

C
125 g

D
27.5 g

Solution

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 =
Question 15
4 / -1

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

A
Volume depends on temperature and molarity involves volume

B
Molarity involves non-volatile solute while all other terms involve volatile solute

C
Number of moles of solute change with change in temperature

D
Molarity is used for polar solvents only

Solution

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.