The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol - 1
∴ Number of moles present in 1000 g of water = 1000/18
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
x2 = 1 / (1+55.56) = 0.0177.
It is given that,
Vapour pressure of water, p10 = 12.3 kPa
Applying the relation, (P10 - P1) / P10 = X2
⇒ (12.3 - p1) / 12.3 =0.0177
⇒ 12.3 - P1 = 0.2177
⇒ p1 = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
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(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone
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(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
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(viii) Aniline to chlorobenzene
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(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
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(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
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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.
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