Question

The elevation in boiling point caused by a dilute solution of diethyl ether is 0.216 K .
Find the relative lowering of vapour pressure of this dilute solution.


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Answer 1

$\tt \red {Given :- }$

The boiling point elevation constant of ether$\bf \: k_{b} = 0.216 \: k \: \: kg/mol$

$\tt \pink{To \:find :-}$

The relative lowering of vapour pressure of this dilute solution.

$\tt \purple{Solution \: :-}$

The molality 'm' of the solution is given by

$\implies \bf m = \frac {w_{2} \times 1000}{ M_{2} \times w_{1} } \: \: \: \: \: \: ..eq(1)$

$\bf \orange{Here,}$

w₁ and w₂ are the masses and

M ₁ and M₂ are the molecular masses of the solvent and solute respectively.

The elevation in boiling point is given as :-

$\implies \bf \: ∆T_{b} = k_{b} \times m \: \: \: \: \: \: \: ..eq(2)$

$\bf \green{Here,}$

∆Tb is the elevation in boiling point and ∆Kb is the boiling point elevation constant.

On Combining equations (1) and (2) ,

$\implies \bf \: ∆T_{b} = k_{b} \times \frac {W_{2} \times 1000}{ M_{2} \times W_{1} } \: \: \: \: \: \: ..eq(3)$

Relative lowering of vapour pressure is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent and is equal to the mole fraction of the solute .

For dilute solutions, the relative lowering of vapour pressure is given by

$\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac {W_{2} \times 1000}{ M_{2} \times W_{1} } \: \: \: \: \: ..eq(4)$

$\bf\blue{Here,}$

$\tt \: p_{s}$ is the vapour pressure of the solution and p⁰ is the vapour pressure of the pure solvent.

$\tt\pink{Therefore,}$

The molecular mass of diethyl ether is 74 g/ mol.

On combining equations (3) and (4),

$\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac{∆T_{b} \times M _{1}}{ K _{b} \times 1000 } \: \: \: \: \: \:.. eq(5)$

SUBSTITUTING THE VALUES IN EQUATION (5),

$\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac{0.216K \times 74 \: g/mol}{2.16K \: kg/mol \: \times 1000 \: g/kg}$

$\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } \: = 0.0074$

HENCE,

The relative lowering of vapour pressure of this dilute solution is 0.0074.

Answer 2

Explanation:

$\tt \red {Given :- }$

The boiling point elevation constant of ether

$\bf \: k_{b} = 0.216 \: k \: \: kg/molk$

=0.216kkg/mol

$\tt \pink{To \:find :-}$

The relative lowering of vapour pressure of this dilute solution.

$\tt \purple{Solution \: :-}$

The molality 'm' of the solution is given by

$\implies \bf m = \frac {w_{2} \times 1000}{ M_{2} \times w_{1} } \: \: \: \: \: \: ..eq \\ (1)⟹m= </p><p>M </p><p>2</p><p> </p><p> ×w </p><p>1</p><p> </p><p> </p><p>w </p><p>2</p><p> </p><p> ×1000</p><p> </p><p> </p><p>$

w₁ and w₂ are the masses and

$\bf \orange{Here,}$

M ₁ and M₂ are the molecular masses of the solvent and solute respectively.

The elevation in boiling point is given as :-

\implies \bf \: ∆T_{b} = k_{b} \times m \: \: \: \: \: \: \: ..eq(2)⟹∆T

b

=k

b

×m..eq(2)

\bf \green{Here,}Here,

∆Tb is the elevation in boiling point and ∆Kb is the boiling point elevation constant.

On Combining equations (1) and (2) ,

$\implies \bf \: ∆T_{b} = k_{b} \times \frac {W_{2} \times 1000}{ M_{2} \times W_{1} } \: \: \: \: \: \: ..eq(3)⟹∆T </p><p>b$

=k

b

×

M

2

×W

1

W

2

×1000

..eq(3)

Relative lowering of vapour pressure is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent and is equal to the mole fraction of the solute .

For dilute solutions, the relative lowering of vapour pressure is given by

\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac {W_{2} \times 1000}{ M_{2} \times W_{1} } \: \: \: \: \: ..eq(4)⟹

p

o

p

o

−p

s

=

M

2

×W

1

W

2

×1000

..eq(4)

\bf\blue{Here,}Here,

\tt \: p_{s}p

s

is the vapour pressure of the solution and p⁰ is the vapour pressure of the pure solvent.

\tt\pink{Therefore,}Therefore,

The molecular mass of diethyl ether is 74 g/ mol.

On combining equations (3) and (4),

\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac{∆T_{b} \times M _{1}}{ K _{b} \times 1000 } \: \: \: \: \: \:.. eq(5)⟹

p

o

p

o

−p

s

=

K

b

×1000

∆T

b

×M

1

..eq(5)

SUBSTITUTING THE VALUES IN EQUATION (5),

\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } = \frac{0.216K \times 74 \: g/mol}{2.16K \: kg/mol \: \times 1000 \: g/kg}⟹

p

o

p

o

−p

s

=

2.16Kkg/mol×1000g/kg

0.216K×74g/mol

\implies\bf \frac{{p}^{o} - p_{s}}{ {p}^{o} } \: = 0.0074⟹

p

o

p

o

−p

s

=0.0074

HENCE,

The relative lowering of vapour pressure of this dilute solution is 0.0074.