Solution of NCERT In Text Questions (Part - 2)
Question – 2.5 - Calculate
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL – 1.
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.
Given, Solubility of H2 S in water at STP = 0.195 m
This means, moles of H2 S = 0.195
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.
Solutions - Class 12 chemisty - NCERT In Text Solution (Part 1)
Solutions - Class 12 chemisty - NCERT In Text Solution (Part 3)
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)