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. K_a = `1/4xx10^(-3)` K_f = 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 P^total, P^chloroform and P^acetone as a function of X^acetone. The experimental data observed for different compositions of mixture is:

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

Answer:

Here, the plot for P_total 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

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