Class 12 Chemistry


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.

Answer: 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

Now, Molality of KI `=(W_B\xx1000)/(M_B\xx\W_A\gm)`

`=(20xx1000)/(166xx80)=(250)/(166)=1/5` m

Molarity of KI `=(W_B\xx\text(Density)xx1000)/(M_B\xx\text(Solution in gm))`


`=(240.4)/(166)=1.4457=1.45` M

Thus, molarity of KI = 1.45 M

Number of moles of KI `=text(Mass of KI)/text(Molecular mass of KI)`

`=(20)/(166)=0.12` mol

Number of moles of H2O `=(text(Mass of H)_2O)/(text(Mole fraction of H)_2O)`

`=(80)/(18)=4/4` mol

Now, Mole fraction of KI `X_(KI)=(n_(KI))/(n_(KI)+n_(H_2O))`


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.

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

This means, moles of H2 S = 0.195

Moles of H2O `=(1000\g)/(18\g\text(mol)^(-1))=55.55` mol

Now mole fraction of H2 (XH2S) `=(text(Mole of H)_2S)/(text(Mole of water)+text(Mole of H)_2S)`


Now, according to Henry’s Law pressure p = KHx

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

Thus, `0.987=K_H\xx0.0035`

Or, `K_H=(0.987)/(0.0035)=282` 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.

Answer: Given Henry’s Law constant `K_H=1.67xx10^8` Pa

PCO2 = 2.5 atm `= 2.5xx101325` Pa

Molar mass of CO2 `=12+16xx2=44` g mol-1

Now, as per Henry’s Law `P_(CO_2)=K_H\xx\X_(CO_2)`

Or, `2.5xx101325Pa=1/67xx10^8Pa\xx\X_(CO_2)`

Or, `X_(CO_2)=(2.5xx101325)/(1.67xx10^8)`



Or, `X_(CO_2)=1.517xx10^(-3)`

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) `=(500)/(18)=27.78` mole

So, `(n_(CO_2))/(n_(H_2O))=X_(CO_2)`

Or, `(n_(CO_2))/(27.78)=1.517xx10^(-3)`

Or, `n_(CO_2)=1.517xx10^(-3)xx27.78`

`=42.14xx10^(-3)` mol

Now, mole of CO2 `=(text(Mass of CO)_2)/(text(Molar mass of CO)_2)`

Or, Mass of CO2 = Molar mass of CO2 `xx` nCO2

Or, Mass of CO2 = 44 g mol-1 `xx` 42.14 `xx` 10-3 mol

`= 1.854 g