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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(ii) Potassium tetracyanonickelate(II)
(iii) Tris(ethane-1,2-diamine) chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
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(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl ketone
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The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = k (PCH3OCH3)3/2
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How will you convert?
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(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
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