Estimate the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at 100°C:a)10 torrb)18 torrc)13.45 torrd)24 torrCorrect answer is option 'C'. Can you explain this answer?

Question

Estimate the lowering of vapour pressure due to the solute (glucose) in a 1.0 M aqueous solution at 100&#176;C:a)10 torrb)18 torrc)13.45 torrd)24 torrCorrect answer is option 'C'. Can you explain this answer?

Accepted Answer

First of all find mole fraction of gas .

Let mole fraction of gas is x

we know, the relation between mole fraction and molality

molality = x x 1000/(1 - x)M

Here, M is Molecular weight of solvent .

for aqueous solution , solvent is water .

∴ Molecular weight of water , M = 18g/mol

Now, 1 = x x 1000/(1 - x) x 18

⇒18(1 - x) = 1000x

⇒18 = (1000 + 18)x

⇒x = 18/1018

Now, use formula ,

Relative lowering of vapor pressure = mole fraction of gas

∆P/P₀ = x

Here P₀ is initial pressure ,at STP , P₀ = 760 torr

so, ∆P = 760 � 18/1018 = 13.45 torr

Hence, lowering of vapor pressure = 13.45 torr

Answer

Lowering of Vapour Pressure

When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases due to the decrease in the number of solvent molecules at the surface. This phenomenon is known as the lowering of vapour pressure. The lowering of vapour pressure is directly proportional to the concentration of the solute in the solution.

Formula for Lowering of Vapour Pressure

The formula for lowering of vapour pressure is given by:

ΔP = P₀ - P₁

where,

ΔP = lowering of vapour pressure

P₀ = vapour pressure of pure solvent

P₁ = vapour pressure of the solution

The lowering of vapour pressure can also be calculated using the following formula:

ΔP = X₂P₀

where,

X₂ = mole fraction of solute in the solution

P₀ = vapour pressure of pure solvent

Calculation of Lowering of Vapour Pressure

In this question, the concentration of the solute (glucose) in the solution is given as 1.0 M and the temperature is given as 100°C.

We can calculate the lowering of vapour pressure using the formula:

ΔP = X₂P₀

where,

X₂ = 1.0 M / (1.0 M + 55.5 M)

= 0.0176

(Here, 55.5 M is the molarity of water)

P₀ = vapour pressure of pure water at 100°C

= 760 torr

ΔP = 0.0176 x 760 torr

= 13.45 torr

Therefore, the correct answer is option C) 13.45 torr.

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