Question

Benzene and naphthalene form ideal solution over the entire range of
composition.The vapour pressures of pure benzene and naphthalene at 300,K are
50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of
benzene in the vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.


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Answer 1

Answer:

0.68

Explanation:

First We Have To Find The Total Pressure

For This We Can Use Dalton's Law

Let Total Pressure Be P

Vapour Pressure of Pure Benzene be P1

Vapour Pressure of Pure Naphthalene be P2

P = P(B) + P(N)

We Know By Roult's Law

P(B) = P1 × Mole Fraction of Benzene

P(N) = P2 × Mole Fraction of Naphthalene

Mole Fraction of Benzene(B) = n(b)/( n(b) + n(n)

Mole Fraction of Naphthalene(N) = n(n)/( n(b) + n(n)

So,

B = 80/78/(80/78 + 100/128)

On Solving

B = 0.57

and By Relation

N + B = 1

N = 1 - 0.57

N = 0.43

Now Putting The Values

P = 50.71 * 0.57 + 32.06*0.43

P = 42.69 mm Hg

Now We Have To Calculate Mole Fraction of Benzene in Vapour Phase

So By Using Formula

Y(B) = P1 × B / P

Putting The Values

Y(B) = 50.71 * 0.57 / 42.69

Y(B) = 0.68

So The Mole Fraction of Benzene in Vapour Phase is 0.68

Answer 2

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Answer:

0.68

Explanation:

First We Have To Find The Total Pressure

For This We Can Use Dalton's Law

Let Total Pressure Be P

Vapour Pressure of Pure Benzene be P1

Vapour Pressure of Pure Naphthalene be P2

P = P(B) + P(N)

We Know By Roult's Law

P(B) = P1 × Mole Fraction of Benzene

P(N) = P2 × Mole Fraction of Naphthalene

Mole Fraction of Benzene(B) = n(b)/( n(b) + n(n)

Mole Fraction of Naphthalene(N) = n(n)/( n(b) + n(n)

So,

B = 80/78/(80/78 + 100/128)

On Solving

B = 0.57

and By Relation

N + B = 1

N = 1 - 0.57

N = 0.43

Now Putting The Values

P = 50.71 * 0.57 + 32.06*0.43

P = 42.69 mm Hg

Now We Have To Calculate Mole Fraction of Benzene in Vapour Phase

So By Using Formula

Y(B) = P1 × B / P

Putting The Values

Y(B) = 50.71 * 0.57 / 42.69

Y(B) = 0.68

So The Mole Fraction of Benzene in Vapour Phase is 0.68