Chemistry Class Twelve

# Solutions

## Solution of NCERT In Text Questions (Part - 2)

Question – 2.5 - Calculate

(a) molality

(b) molarity and

(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL – 1.

Solution:

Given, 20% (mass/mass) aqueous solution of KI.

This means 20 gm of KI is dissolved in 80 gm of water.

Molar mass of KI = 39 + 127 = 166 g mol – 1

Mass of solute (W B) = 20 gm

Mass of solvent (WA ) = 80 gm

Molar mass of solute (M B)= 166 g mol – 1

Density of the aqueous KI = 1.202 g mL – 1

(a) Now, Molality of KI

(b) Molarity of KI

Thus, molarity of KI = 1.45 M

(c) Number of moles of KI

Number of moles of H2O

Now, Mole fraction of KI

Thus, mole fraction of KI in the given solution = 0.265

Question – 2.6 - H 2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2 S in water at STP is 0.195 m, calculate Henry’s law constant.

Solution:

Given, Solubility of H2 S in water at STP = 0.195 m

This means, moles of H2 S = 0.195

Moles of

Now mole fraction of

Now, According to Henry’s Law

We, know that, pressure (p) at STP = 0.987 bar

Thus, Henry’s Law constant KH = 282 bar

Question – 2.7 - Henry’s law constant for CO2 in water is 1.67x108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO 2 pressure at 298 K.

Solution:

Given,

Henry’s Law constant

Molar mass of

Now, As per Henry’s Law

Now, number of mole of water present in 500 mL of soda water

Molar mass of water (H 2O) = 1 x 2 + 16 = 18

Mass of water = 500 mL = 500 g (since density of water = 1 g m3)

Therefore, number of moles of water (H 2O)

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Solutions - Class 12 chemisty - NCERT In Text Solution (Part 1)
Solutions - Class 12 chemisty - NCERT In Text Solution (Part 3)

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