A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
1) Molar mass of the solute
2) Vapour pressure of water at 298 K.
Let, the molar mass of the solute be M g mol - 1
Now, the no. of moles of solvent (water),n1 = 90g / 18g mol-1
And, the no. of moles of solute,n2 = 30g / M mol-1 = 30 / M mol
p1 = 2.8 kPa
Applying the relation:
(p10 - p1) / p10 = n2 / (n1 + n2)
⇒ (p10 - 2.8) / p10 = (30/M) / {5 + (30/M)}
⇒ 1 - (2.8/p10) = (30/M) / {(5M+30)/M}
⇒ 1 - (2.8/p10) = 30 / (5M + 30)
⇒ 2.8/p10 = 1 - 30 / (5M + 30)
⇒ 2.8/p10 = (5M + 30 - 30) / (5M + 30)
⇒ 2.8/p10 = 5M / (5M+30)
⇒ p10 / 2.8 = (5M+30) / 5M ----------------(1)
After the addition of 18 g of water:
n1 = (90+18g) / 18 = 6 mol
and the new vapour pressure is p1 = 2.9 kPa (Given)
Again, applying the relation:
(p10 - p1) / p10 = n2 / (n1 + n2)
⇒ (p10 - 2.9) / p10 = (30/M) / {6 + (30/M)}
⇒ 1 - (2.9/p10) = (30/M) / {(6M+30)/M}
⇒ 1 - (2.9/p10) = 30 / (6M + 30)
⇒ 2.9/p10 = 1 - 30 / (6M + 30)
⇒ 2.9/p10 = (6M + 30 - 30) / (6M + 30)
⇒ 2.9/p10 = 6M / (6M+30)
⇒ p10 / 2.9 = (6M+30) / 6M ----------------(2)
Dividing equation (1) by (2),we get:
2.9 / 2.8 = {(5M+30) / 5M} / {(6M+30) / 6M}
⇒ 2.9 x (6M+30 / 6) = (5M+30 / 5) x 2.8
⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23u
Therefore, the molar mass of the solute is 23 g mol - 1.
(ii) Putting the value of 'M' in equation (i), we get:
⇒ p10 / 2.8 = (5M+30) / 5M
⇒ p10 / 2.8 = (5x23+30) / 5x23
⇒ p10 = (145 x 2.8) / 115
⇒ p10 = 3.53
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
Give simple chemical tests to distinguish between the following pairs of compounds.
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
(vii) Ethanal and Propanal
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL-1, then what shall be the molarity of the solution?
Henry's law constant for CO2 in water is 1.67 x 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K.
Calculate the mass of a non-volatile solute (molar mass 40 g mol-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?
Distinguish between
(i)Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3-
(ii) [PtCl2(en)2]2+
(iii) [Cr(NH3)2Cl2(en)]+
Sea is the greatest source of some halogens. Comment.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Explain the term copolymerisation and give two examples.
What are ambident nucleophiles? Explain with an example.
What are biodegradable and non-biodegradable detergents? Give one example of each.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
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 Kif the Henry's law constants for oxygen and nitrogen are 3.30 x 107 mm and 6.51 x 107mm respectively, calculate the composition of these gases in water.
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Why are you adding 1 in p°-p/p° ??
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