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3 years ago

A mixture of He

Chemistry

1 answer:

aksik [14]3 years ago

3 0

Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm

Explanation :

According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.

Formula used :

$p_T=p_{He}+p_{Ar}+p_{N_2}$

where,

$p_T$ = total pressure of gas = 13.6 atm

$p_{He}$ = partial pressure of helium gas = 1831 torr = 2.41 atm

$p_{Ar}$ = partial pressure of argon gas = 997 torr = 1.31 atm

Conversion used: (1 atm = 760 torr)

$p_{N_2}$ = partial pressure of nitrogen gas = ?

Now put all the given values in the above formula, we get:

$13.6=2.41+1.31+p_{N_2}$

$p_{N_2}=9.88atm$

Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm

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The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:

p₁ = χ₁p₁°

where

χ₁ = the mole fraction of the solvent

p₁ and p₁° are the vapour pressures of the solution and of the pure solvent

The formula for vapour pressure lowering Δp is

Δp = p₁° - p₁

Δp = p₁° - χ₁p₁° = p₁°(1 – χ₁) = χ₂p₁°

where χ₂ is the mole fraction of the solute.

Step 1. Calculate the mole fraction of glucose

n₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu

n₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O

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Step 2. Calculate the vapour pressure lowering

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr

Step 3. Calculate the vapour pressure

p₁ = p₁° - Δp = 23.8 torr – 0.4430 torr = 23.4 torr

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3 years ago