Class 12 Chemistry

Solutions

NCERT Solution Part 4

Question 32: Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = `1/4xx10^(-3)` Kf = 1.86 K kg mol-1.

Answer: Molar mass of CH3CH2CHClCOOH

`= 4xx12+7xx1+35.5xx2x16=48+7+35.5+32=122.5` g mol-1

Number of moles in 10 g of given acid `=(10)/(122.5)=8.16xx10^(-2)` mole

Molality `=(8.16xx10^(-1))/(250)xx1000=0.3264` m

If α is the degree of dissociation of given acid and C is the initial concentration

CH3CH2CHClCOOH ⇄ CH3CH2CHClCOO- + H+

Initial concentration = C → 0 + 0

Concentration at equilibrium = C(1 - α) ⇄ Cα + Cα

So, `K_α=(C^2α^2)/(C(1-α))`

`=(Cα^2)/(1-α)=Cα^2`

Because α is very small

Or, `α=sqrt((K_a)/C)`

`=sqrt((1.4xx10^(-3))/(0.3264))=0.065`

Calculation of van’t Hoff factor:

Initially the concentration of given acid will be 1 mole

So, `i=(1+α)/1=1+α`

`=1+0.065=1.065`

So, `ΔT_f=i\K_f\m`

`=1.065xx1.86xx0.3264=0.65°` K


Question 33: 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0° C. Calculate the van;t Hoff factor and dissociation constant of fluoroacetic acid.

Answer: We know:

`M_2=(1000K_f\W_2)/(W_1ΔT_f)`

So, Observed `M_2=(1000xx1.86xx19.5)/(500xx1)=72.54` g mol-1

M2 (Calculated) `=12+2+19+12+2xx16+1=78` g mol-1

Hence, `i=(M_2text([calculated]))/(M_2text([observed]))`

`=(78)/(72.54)=1.0753`

Calculation of dissociation constant:

Let us assume that α is the degree of dissociation and C is the initial concentration

`i=(C(1+α))/C=1+α`

Or, `α=i-1`

`=1.0753-1=0.0753`

`K_α=(0.5xx(0.0753)^2)/(1-0.0753)=3.07xx18^(-3)`

Question 34: Vapor pressure of water at 293 K is 17.535 mm Hg. Calculate the vapor pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer: Given P0 = 17.535 mm Hg

Molar mass of glucose = 180 g mol-1

Molar mass of water = 18 g mol-1

According to Raoult’s Law:

`(P^o-P_s)/(P^o)=(n_2)/(n_1+n_2)`

`=(n_2)/(n_1)=((W_2)/(M_2))/((W_1)/(M_1))`

So, `1-(P_s)/(P^o)=((25)/(180))/((450)/(18))`

`=(25xx18)/(180xx450)=0.0055`

Or, `(P_s)/(P^o)=1-0.0055=0.9945`

Or, `(P_s)/(17.535)=0.9945`

Or, `P_s=0.9945xx17.535=17.44` mm Hg


Question 35: Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 `xx` 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Answer: We know, `P=K_H\x`

So, `x=P/(K_H)`

`=(760 mm Hg)/(4.27xx10^5 mm Hg)=1.78xx10^(-3)`

This is the mole fraction of methane in benzene

Question 36: 100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1). The vapor pressure of pure liquid B was found to be 500 torr. Calculate the vapor pressure of pure liquid A and its vapor pressure in the solution if the total vapor pressure of the solution is 475 torr.

Answer: Number of moles of solute `n_2=(100)/(140)=5/7` mole

Number of moles of solvent `n_1=(1000)/(180)=(50)/9` mole

Mole fraction of solute `x_2=(n_2)/(n_1+n_2)`

`=(5/7)/(5/7+(50)/9)=0.114`

Mole fraction of solvent `x_1=1-x_2=1-0.114=0.886`

According to Raoult’s Law

`P_A=x_A\P_A^o=0.114xx\P_A^o`

`P_B=x_B\P_B^o`

`=0.886xx500=443` torr

`P_(total)=P_A+P_B`

Or, `475=0.114P_A^o+443`

Or, `P_A^o=(475-443)/(0.114)=280.7` torr

Or, `P_A=0.114xx280.7=32` torr

Question 37: Vapor pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal, Pchloroform and Pacetone as a function of Xacetone. The experimental data observed for different compositions of mixture is:

100 `xx` xacetonePacetone/mm HgPchloroform/mmHg
00632.8
11.854.9548.1
23.4110.1469.4
36.0202.4359.7
50.8322.7257.7
58.2405.9193.6
64.5454.1161.2
72.1521.1120.7

Plot this data also on the same graph paper. Indicate whether it has positive deviation of negative deviation from the ideal solution.

Answer:

100 `xx` xacetonePacetone/mm HgPchloroform/mmHgPTotal
00632.8632.8
11.854.9548.1632.8
23.4110.1469.4579.5
36.0202.4359.7562.1
50.8322.7257.7580.4
58.2405.9193.6599.5
64.5454.1161.2615.3
72.1521.1120.7641.8

graph for deviation from ideal solution

Here, the plot for Ptotal is sloping downward. Hence, the solution shows negative deviation from ideal behavior.

Question 38: Benzene and toluene form ideal solution over the entire range of composition. The vapor pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapor phase if 80 g of benzene is mixed with 100 g of toluene.

Answer: Molar mass of `C_6H_6` = 78 g mol-1

Molar mass of `C_6H_5CH_3` = 92 g mol-1

Number of moles of `C_6H_6=(80)/(78)=1.026` mole

Number of moles of `C_6H_5CH_3=(100)/(92)=1.087` mole

Mole fraction of `C_6H_6`

`x_B=(1.026)/(1.026+1.087)=0.486`


Question 39: The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 `xx` 107 mm and 6.51 `xx` 107 mm respectively, calculate the composition of these gases in water.

Answer: Air contains 20% oxygen by volume

Hence, partial pressure of O2

`P_(O_2)=(20)/(100)xx10=2` atm

`=2xx760=1520` mm

Similarly, partial pressure of N2

`P_(N_2)=(79)/(100)xx10=7.9` atm

`=7.9xx760=6004` mm

According to Henry’s Law

`P_(O_2)=K_H\x_(O_2)`

Or, `x_(O_2)=(1520)/(330xx10^7)=4.61xx10^(-5)`

Similarly,

`x_(N_2)=(6004)/(6.51xx10^7)=9.22xx10^(-5)`

Question 40: Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 at 27° C.

Answer: We know that:

`π=i\C\R\T=(i\n)/(V\RT)`

Or, `n=(πV)/(i\RT)`

`=(0.75xx2.5)/(2.47xx0.0821xx300)=0.0308` mole

Molar mass of `CaCl_2=40+2xx35.5=111` g mol-1

Hence, amount of `CaCl_2` dissolved

`=0.0308xx111=3.42` g

Question 41: Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25° C, assuming that it is completely dissociated.

Answer: Molar mass of `K_2SO_4=2xx39+32+4xx16`

`=78+32+64=174` g mol-1

Dissociation of `K_2SO_4` can be given by following equation:

`K_2SO_4 ↠2K^+\+SO_4^(2-)`

This shows that number of ions produced = 3, i.e. `i=3`

Using the equation `π=i\C\R\T=(i\n)/(V\RT)`

Or, `π=i\xx\W/M\xx(RT)/V`

`=(3xx0.025xx0.0821xx298)/(174xx2)=5.27xx10^(-3)` atm



Molality

InText Questions

NCERT Solution 1

NCERT Solution 2

NCERT Solution 3

NCERT Solution 4