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
Because α is very small
Calculation of van’t Hoff factor:
Initially the concentration of given acid will be 1 mole
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:
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
Calculation of dissociation constant:
Let us assume that α is the degree of dissociation and C is the initial concentration
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:
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`
`=(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)`
Mole fraction of solvent `x_1=1-x_2=1-0.114=0.886`
According to Raoult’s Law
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|
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|
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`
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
Similarly, partial pressure of N2
According to Henry’s Law
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:
Molar mass of `CaCl_2=40+2xx35.5=111` g mol-1
Hence, amount of `CaCl_2` dissolved
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:
This shows that number of ions produced = 3, i.e. `i=3`
Using the equation `π=i\C\R\T=(i\n)/(V\RT)`
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