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QUESTION: 1

**Only One Option Correct Type
This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct**

Which of the following concentration factors is affected by change in temperature?

Solution:

On changing temperature, solvent (H_{2}O) is vaporised or condensed. Thus, volume of solution and solvent changes. Thus, any concentration factor dependent on volume is affected by change in temperature.

(a) Molarity = Mole/Litre (Volume dependent)

(b) Molality = Moles of solute/kg of solvent (weight dependent)

(c) Mole fraction = n_{i}/n_{total}(unitless)

(d) Weight fraction = w_{i}/w_{total} (unitless)

QUESTION: 2

6.02 x10^{20} molecules of urea are present in 100 mL of its solution. The concentration of urea solution is (N_{0} = 6.02 x 10^{23} mol^{-1})

[AIEEE 2004]

Solution:

QUESTION: 3

Glucose solution is one molal. Glucose present in 1 kg glucose solution is

Solution:

Glucose solution is one molal. Thus, 1000 g water (solvent) has glucose = one mole = 180 g

Total mass of solution = 1000 +180 = 1180 g

1180 g of solution has glucose = 180 g

∴ 1000 g of solution has glucose =

180 x 1000/1180 = 152.5 ≈ 153g

QUESTION: 4

The molality of a urea solution in which 0.0100 g of urea (NH_{2}CONH_{2}) is added to 0.300 dm^{3} of water at STP is

[AIEEE 2011]

Solution:

QUESTION: 5

A solution that is 20% ethanol by volume is found to have a density of 0.977 g/mL. Density of ethanol is 0.789 g/mL. Thus, mass per cent of ethanol solution is

Solution:

20% ethanol by volume means 100 m l of (ethanol + H_{2}0 ) solution has ethanol = 20 mL 100 mL of solution = 100 x 0.977 g solution = 97.7 g solution 20 mL ethanol = 20 x 0.789 g ethanol = 15.78 g

QUESTION: 6

A solution of NH_{4}CI is prepared by dissolving 95 g of NH_{4}CI in 200 g of H_{2}O at 60° C. What is mass per cent when the solution is cooled to 20° C based on this

Solution:

200 g H_{2}O at 60°C has 95 g NH_{4}Cl

Based on solubility graph, at 20°C,

100 g of H_{2}O can dissolve = 36 g NH_{4}CI

hence, 200 g of H_{2}0 can dissolve = 72g NH_{4}CI

Thus, NH_{4}CI separated (crystallised) = 95 - 72 = 23 g

Mass % = 36%

QUESTION: 7

Commercial concentrated nitric acid is 15.6 M. To prepare 10 L of 6.0 M nitric acid from it,

Solution:

This is a case of dilution.

Thus, 3.846 L of 15.6 M solution is diluted to 10.0 L by addition of 6.154 L of H_{2}0.

QUESTION: 8

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

Solution:

The correct answer is option C

The molar mass M_{2} of non-volatile substance can be calculated from the following formula.

Here, p^{0} vapour pressure of acetone and p is the vapour pressure of the solution.

M_{1} and W_{1} are the molar mass and mass of acetone.

M_{2} and W_{2} are the molar mass and mass of a non-volatile substance.

Substitute values in the above expression.

Hence, M2=64.38g/mol.

Thus, the molar mass of the non-volatile substance is 64 g/mol (approximately).

QUESTION: 9

AgNO_{3} sample is 85% by mass. To prepare 125 mL of 0.05 M AgNO_{3} solution, AgNO_{3} sample required is

Solution:

125 mL of 0.05 M AgN0_{3} = 125 x 0.05 millimoles

QUESTION: 10

Hardness of a water sample is 200 ppm CaCO_{3}, Thus, molarity of CaCO_{3} is

Solution:

Hard water has hardness = 200 ppm Thus, 10^{6} mL(= 10^{3} L)water has CaC0^{3} =200 g =2 mol

QUESTION: 11

100 mL of aqueous solution of 0.01 M CaCI_{2} is evaporated to dryness when 0.15 g of residue is obtained. Thus, impurity present is

Solution:

100mLof 0.01 M CaCI_{2} = 100 x 0.01 millimoles CaCI_{2}

= 0.001 mole CaCI_{2}

= 0.001 x 111 g CaCI_{2}

= 0.111 g pure CaCI_{2}

But, residue = 0.15 g

Thus, impurity = 0.15 - 0.111= 0.039 g

QUESTION: 12

At 25 ° C, the density of 18 MH_{2}SO_{4} is 1.8 g cm^{3}. Thus, mass percentage of H_{2}SO_{4} in aqueous solution is

Solution:

18 MH_{2}SO_{4} means18 mol L^{-1}.

1000 mL of H_{2}S0_{4} solution has H_{2}SO_{4} = 18 mol

= 18 x98g

Also, 1000 m L = 1000 x 1.8 g H_{2}SO_{4} solution

Thus, (1000 x 1.8)g H_{2}SO_{4} solution has H_{2}SO_{4} =18 x 98 g

QUESTION: 13

Laboratory ammonia is 14.8 M NH_{3}(ag)with a density of 0.8980 g/mL. Mole fraction of NH_{3} in this solution is

Solution:

14.8 M NH_{3}(ag ) means 14.8 moles NH_{3} in 1000 mL solution . 1000 mL solution = 1000 x 0,898 g solution

QUESTION: 14

An aqueous solution is 34% H_{3}P0_{4} by mass and has density 1.209 g mL ^{-1} Molarity (I), molality (II) and normality (III) respectively, are

Solution:

100g H_{3}P0_{4} aqueous solution has 34g H_{3}P0_{4}

QUESTION: 15

If one assumes the volumes are additive, what is in a solution obtained by mixing 275 mL of 0.200 M KN0_{3}, 325 mL of 0.40 M Mg(N0_{3})_{2} and 400 mL H_{2}0?

Solution:

QUESTION: 16

How many grams of water would you add to 1.38 moles of CH_{3}OH in 1 kg water to reduce the molality to 1.00 molal CH_{3}OH(aq) ?

Solution:

1.38 moles of CH_{3}OH in w kg water

1.38 moles of CH_{3}OH in w Kg water

So that molarity = 1.0

QUESTION: 17

Three solutions have been provided.

I. 3 g HCI in 1 kg H_{2}0.

II. 2 L HCI(g) and 1 L H_{2}0 at room temperature.

III. Aqueous solution of HCI with

Solutions containing same number of moles of HCI is/are

Solution:

QUESTION: 18

Which of the following aqueous solutions has the highest concentration of Na^{+}?

Solution:

QUESTION: 19

pH of Ba(OH)_{2} solution is 12. Number of millimoles present in 100 mL of Ba(OH)_{2} solution is

Solution:

QUESTION: 20

We have

I. 25 mL of 1 M NaOH

II. 10 mL of 0.50 M NaCI

On mixing the two solutions, molar concentrations of Na+, OH^{-} and Cl^{-} respectively, are

Solution:

This is a case of dilution.

NaOH solution has been diluted by NaCI or vice-versa.

Total volume = 35.0 mL

M_{1}V_{1} = M_{2}V_{2}

QUESTION: 21

A ‘100 proof solution of ethanol in water consists of 50.0 mL of C_{2}H_{5}OH(/)and 50.0 mL of H_{2}0 (/), mixed at 16 °C . Given,

Density of H_{2}0 = 1 g mL^{-1}

Density of C_{2}H,OH = 0.7939 gmL^{-1}

Density mixture = 0.9344 g mL^{-1}

Volume of the solution is

Solution:

Volume can be additive or not depend on the nature of the solution

QUESTION: 22

How many grams of Mgl_{2} must be added to 250 mL of 0.0876 M Kl to produce a solution with [l^{-}] = 0.10 M?

Solution:

QUESTION: 23

An aqueous solution is 6% methanol, CH_{3}OH, by mass with d = 0.988 g mL^{-1}. Thus, molarity of CH_{3}OH in this solution is

Solution:

(d) Aqueous 6% methanol solution by mass means 100 g of aqueous solution has = 6g CH_{3}OH

QUESTION: 24

**Matching List Type
Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct**

**An ethanol-water solution is prepared by dissolving 10.00 mL o f ethanol (C _{2}H_{5}OH, d = 0.789 g mL^{-1}) in a sufficient volume o f w ater to produce 100 mL of solution with a density, d = 0.982 g mL^{-1}. Match the concentration term in Column I with its value in Column II and select the answer from the code**

Solution:

*Multiple options can be correct

QUESTION: 25

**One or More than One Options Correct Type
This section contains 2 multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.**

1 kg of aqueous solution has 0.40 kg NaOH. Thus, this solution is

Solution:

1 kg aqueous solution = 1000 g aqueous solution

It has 0.4 kg NaOH = 0.4 x1000= 400 g NaOH

Thus, solvent (water) =1000 - 400 = 600

(d) Since, density is not given hence, molarity cannot be derive

*Multiple options can be correct

QUESTION: 26

Mole fraction of ethanol in ethanol-water solution is 0.25. Thus, this solution is

Solution:

QUESTION: 27

**Comprehension Type**

**This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)**

**Passage**

**A handbook gives the concentration of water in a 2.772 M aqueous solution of NaOH as 998.0 g L ^{-1}.**

Q. Molality of solution is

Solution:

Molarity = 2.772 M =2.772 mol L^{-1 }Thus, NaOH =2.772 mol

QUESTION: 28

**Comprehension Type**

**This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)**

**Passage**

**A handbook gives the concentration of water in a 2.772 M aqueous solution of NaOH as 998.0 g L ^{-1}.**

**Q.**

Molality of solution is

Solution:

Molarity = 2.772 M =2.772 mol L^{-1 }Thus, NaOH =2.772 mol

*Answer can only contain numeric values

QUESTION: 29

**One Integer Value Correct Type
This section contains 1 questions, when worked out will result in an integer value from 0 to 9 (both inclusive)**

Q. Certain brine solution has 3.87% NaCI by mass with density 1.036 g mL^{-1}. How many millilitre of this solution should be evaporated to obtain 0.32 g of NaCI?

Solution:

100 g brine solution has 3.87 g NaCI.

QUESTION: 30

A 5.2 molal aqueous solution of methyl alcohol (CH_{3}OH) is supplied. What is the mole fraction of methyl alcohol in the solution?

Solution:

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