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Test: Strength of Solutions - NEET MCQ


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10 Questions MCQ Test Chemistry Class 12 - Test: Strength of Solutions

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

Molality is expressed in

Detailed Solution for Test: Strength of Solutions - Question 1

Molality = moles of solute / kg of solvent.

Test: Strength of Solutions - Question 2

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?

Detailed Solution for Test: Strength of Solutions - Question 2

1 ppm is equivalent to 1 part out of 1 million (106) parts.
∴ Mass percent of 15 ppm chloroform in water 

⇒ 1.5 x 10-3 g chloroform present in 100 g water
Thus, 15g chloroform will be present in water 

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Test: Strength of Solutions - Question 3

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

Detailed Solution for Test: Strength of Solutions - Question 3

Molecular mass of glycerine = 92 g
Molecular mass of water = 18 g

No. of moles of glycerine = 46/92 = 0.5 moles
No. of moles of water = 36/18 = 2 moles

Thus, Mole fraction of glycerine = No. of moles of glycerine/(No. of moles of glycerine + No. of moles of water) = 0.5/(2 + 0.5)  = 0.20

Test: Strength of Solutions - Question 4

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

Detailed Solution for Test: Strength of Solutions - Question 4

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

⇒ molality = 0.127 x 1000/87.5 = 1.45 m

Test: Strength of Solutions - Question 5

When the solvent is in solid state, solution is

Detailed Solution for Test: Strength of Solutions - Question 5

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

Solid Sol includes gems, photochromic glass, alloys, ruby stone, coloured glass, vision glasses, etc.

Test: Strength of Solutions - Question 6

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

Detailed Solution for Test: Strength of Solutions - Question 6

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

Test: Strength of Solutions - Question 7

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

Detailed Solution for Test: Strength of Solutions - Question 7

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

10= 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

Test: Strength of Solutions - Question 8

What is the mass of NaCl required to prepare 0.5 liters of a 2.5 molar solution of NaCl?

Detailed Solution for Test: Strength of Solutions - Question 8

The molecular weight of NaCl is 58.44 g/mol

So, one mole of NaCl weighs 58.44 g.

A 2.5 M solution is 2.5 moles per liter (Molarity is just the number of moles per liter).

Therefore, 0.5 L would contain 1.25 mol. Hence, you would need 1.25 × 58.44 g = 73 g.

Test: Strength of Solutions - Question 9

The molarity of a solution obtained by mixing 750 ml of 0.5(M) HCl with 250ml of 2(M) HCl will be?

Detailed Solution for Test: Strength of Solutions - Question 9

The molarity of a resulting solution is given by

Test: Strength of Solutions - Question 10

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.

Detailed Solution for Test: Strength of Solutions - Question 10

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.

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