Test: Strength Of Solutions

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This mock test of Test: Strength Of Solutions for Class 12 helps you for every Class 12 entrance exam. This contains 10 Multiple Choice Questions for Class 12 Test: Strength Of Solutions (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Strength Of Solutions quiz give you a good mix of easy questions and tough questions. Class 12 students definitely take this Test: Strength Of Solutions exercise for a better result in the exam. You can find other Test: Strength Of Solutions extra questions, long questions & short questions for Class 12 on EduRev as well by searching above.

QUESTION: 1

Mole fraction of ethyl chloride and methanol in a ternary solution is 0.6 and 0.32 respectively. What is the mole fraction of third component. Also identify the solvent in this ternary solution.

A.
0.03, Ethyl chloride
B.
0.08, cannot be determined
C.
0.03, Methanol
D.
0.08, Ethyl chloride

Solution:

Mole fraction of third component is 1 - (0.6 + 0.32) = 0.08. Since ethyl chloride has highest mole fraction so it is the solvent.

QUESTION: 2

Molality is expressed in

A.
mol/kg
B.
mol/L
C.
L/mol
D.
g/L

Solution:

Molality = moles of solute/kg of solvent.

QUESTION: 3

The level of contamination of chloroform was found to be 15 ppm. It means 15 g of chloroform is present in how many grams of solution?

A.
1000 g
B.
10⁶ g
C.
100 g
D.
1.0 g

Solution:

1 ppm is equivalent to 1 part out of 1 million (10⁶) parts.

∴ Mass percent of 15 ppm chloroform in water

QUESTION: 4

Mole fraction of glycerine, C₃H₅(OH)₃ in a solution containing 36 gm of water and 46 gm of glycerine is:

A.
0.40
B.
0.20
C.
0.46
D.
0.36

Solution:

No. of moles of glycerine= 46/92 (where 92 is M.M. of glycerine)

= 0.5moles

no. of moles of water= 36/18(where 18 is M.M. of water)

= 2moles

so mole fraction oh glycerine = No. of moles of glycerine/No. of moles of glycerine + No. of moles of water

= 0.5/2 + 0.5 = 0.20

QUESTION: 5

Calculate the molality of 12.5% w/w sulphuric acid?

A.
2.85m
B.
3.15m
C.
1.45m
D.
2.50m

Solution:

12.5% w/w means 12.5 g in 100 g of solution.

Weight of solvent = 100 g – 12.5 g = 87.5 g.
Number of moles of sulphuric acid = 12.5/ 98 = 0.127 mol

So molality = 0.127 x 1000/87.5 = 1.45 m

QUESTION: 6

When the solvent is in solid state, solution is

A.
Solid solution
B.
Gaseous solution
C.
Solution
D.
Liquid solution

Solution:

When both the solute and solvent are in solid state, then the solution is called as solid solution or solid sol.

QUESTION: 7

The number of moles of KCl in 3 L of 3 M solution is

A.
27 moles
B.
1 moles
C.
9 moles
D.
3 moles

Solution:

In 3M there will be 3 moles per litre therefore in 3 litres there will be 9 moles.

QUESTION: 8

A sample of 300.0 g of drinking water is found to contain 38 mg Pb. What this concentration in parts per million?

A.
3 x 10² ppm
B.
6.5 m
C.
130 ppm Pb
D.
21 ppm

Solution:

We use parts per million to express the concentrations of solutions that contain very, very small amounts, often called trace amounts, of a given solute.

More specifically, a solution's concentration in parts per millions tells you the number of parts of solute present for every

106=1,000,000

parts of solution. You can thus say that a 1 ppm solution will contain exactly 1 g of solute for every 10⁶g of solution.

In your case, you know that you have

in exactly
300.0 g = 3.000 ⋅ 10²g solution

This means that you can use this known composition as a conversion factor to scale up the mass of the solution to 10⁶g

Since this represents the mass of lead present in exactly 10⁶g of solution, you can say that the solution has a concentration of