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.
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
I | 0.1 | 0.1 |
6.0 × 10 - 3 |
II | 0.3 | 0.2 |
7.2 × 10 - 2 |
III | 0.3 | 0.4 |
2.88 × 10 - 1 |
IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
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