Test: Types and Solubility of Solutions (October 17) - NEET MCQ

Test: Types and Solubility of Solutions (October 17) - NEET MCQ

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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Types and Solubility of Solutions (October 17)

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Test: Types and Solubility of Solutions (October 17) - Question 1

During dissolution when solute is added to the solvent, some solute particles separate out from the solution as a result of crystallisation. At the stage of equilibrium, the concentration of solute in the solution at given temperature and pressure

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 1

At dynamic equilibrium, number of solute particles going into the solution will be equal to solute particles separating out. Hence, the concentration of solute in the solution remains constant.

Test: Types and Solubility of Solutions (October 17) - Question 2

According to Henry's law the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution. For different gases the correct statement about Henry's constant is

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 2

p = KHx. Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid.

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Test: Types and Solubility of Solutions (October 17) - Question 3

H2S is a toxic gas used in qualitative analysis. If solubility of H2S in water at STPSTP is 0.195m, what is the value of KH?

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 3

No. of moles of H2S = 0.195
No. of moles of H2O = 1000/18 = 55.55mol
Mole fraction of H2S =
Pressure at STP = 0.987 bar
According to Henry’s law, p = KHx
or KH = pH2S/xH2S = 0.98/70.0035
= 282 bar

Test: Types and Solubility of Solutions (October 17) - Question 4

When a gas is bubbled through water at 298K, a very dilute solution of gas is obtained. Henry’s law constant for the gas is 100kbar. If gas exerts a pressure of 1bar1bar, the number of moles of gas dissolved in 11 litre of water is

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 4

p = KH × x

Mole fraction = Moles of gas/Total moles
Moles of H2O = 1000/18 = 55.55 (∵ 1L = 1000 g)
Mole fraction = (55.55 >>> x)

Test: Types and Solubility of Solutions (October 17) - Question 5

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

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 5

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: Types and Solubility of Solutions (October 17) - Question 6

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

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 6

ppm =

= 10.4 ppm

Test: Types and Solubility of Solutions (October 17) - Question 7

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

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 7

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

Test: Types and Solubility of Solutions (October 17) - Question 8

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

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 8

No. of moles of NaOH

Mass of NaOH = 40 × 0.125 = 5g

Test: Types and Solubility of Solutions (October 17) - Question 9

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: Types and Solubility of Solutions (October 17) - Question 9

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: Types and Solubility of Solutions (October 17) - Question 10

The molality of 648 g of pure water is

Detailed Solution for Test: Types and Solubility of Solutions (October 17) - Question 10

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

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