Directions : Read the passage given below and answer the following que...
Vapour pressure of water, p1° = 17.535 mm of Hg
Mass of glucose, w2 = 25 g
Mass of water, w1 = 450 g
We know that,
Molar mass of glucose (C6H12O6),
M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
Molar mass of water, M1 = 18 g mol- 1
Then, number of moles of glucose, n1 = 25/180 = 0.139 mol
And, number of moles of water, n2 =450/18 = 25 mol
Now, we know that,
(p1° - p°) / p1° = n1 / n2 + n1
⇒ 17.535 - p° / 17.535 = 0.139 / (0.139+25)
⇒ 17.535 - p1 = 0.097
⇒ p1 = 17.44 mm of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.
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Directions : Read the passage given below and answer the following que...
Colligative Properties and Relative Lowering in Vapour Pressure
- Colligative properties are properties of solutions that depend only on the number of solute particles and not on the nature of the solute.
- Relative lowering in vapour pressure is an example of a colligative property.
- In this experiment, the lowering in vapour pressure of a sugar solution was found to be 0.061 mm of Hg.
Calculating Vapour Pressure of Water with Glucose Solution
- To calculate the vapour pressure of water with glucose solution, we need to use the formula for relative lowering in vapour pressure.
- The formula is:
ΔP/P₀ = (n₂/n₁)
Where ΔP is the lowering in vapour pressure, P₀ is the vapour pressure of the pure solvent (water), n₂ is the number of particles of solute (glucose) and n₁ is the number of particles of solvent (water).
- We are given that the vapour pressure of water at 20°C is 17.5 mm of Hg and that 25g of glucose is dissolved in 450g of water.
- First, we need to calculate the number of particles of glucose and water.
- To do this, we use the formula:
n = m/M
Where n is the number of particles, m is the mass of the substance and M is the molar mass.
- The molar mass of glucose is 180 g/mol.
- Therefore, the number of particles of glucose is:
n₂ = 25g / 180 g/mol = 0.139 mol
- The molar mass of water is 18 g/mol.
- Therefore, the number of particles of water is:
n₁ = 450g / 18 g/mol = 25 mol
- Now, we can substitute these values into the formula for relative lowering in vapour pressure:
ΔP/P₀ = (n₂/n₁)
ΔP/17.5 = (0.139/25)
ΔP = 0.077 mm of Hg
- Finally, we can calculate the vapour pressure of water with glucose solution:
P = P₀ - ΔP
P = 17.5 - 0.077
P = 17.423 mm of Hg
- Therefore, the answer is option B, 17.4 mm of Hg.