# 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.

= 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.

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 xacetone Pacetone/mm Hg Pchloroform/mmHg 0 0 632.8 11.8 54.9 548.1 23.4 110.1 469.4 36.0 202.4 359.7 50.8 322.7 257.7 58.2 405.9 193.6 64.5 454.1 161.2 72.1 521.1 120.7

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

 100 xx xacetone Pacetone/mm Hg Pchloroform/mmHg PTotal 0 0 632.8 632.8 11.8 54.9 548.1 632.8 23.4 110.1 469.4 579.5 36.0 202.4 359.7 562.1 50.8 322.7 257.7 580.4 58.2 405.9 193.6 599.5 64.5 454.1 161.2 615.3 72.1 521.1 120.7 641.8 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.

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