Question

A mixture of ethyl acetate vapour and air has a relative saturation of 50% at 303 K and a total pressure of 100 kPa. If the vapour pressure of ethyl acetate at 303 K is 16 kPa, the molal saturation is (a) 0.080 (b) 0.087 (c) 0.264 (d) 0.029

Answer 1

Answer : The correct option is, (b) 0.087

Explanation :

The formula used for relative saturation is:

$\text{Relative saturation}=\frac{P_A}{P_A^o}$

where,

$P_A$ = partial pressure of ethyl acetate

$P_A^o$ = vapor pressure of ethyl acetate

Given:

Relative saturation = 50 % = 0.5

Vapor pressure of ethyl acetate = 16 kPa

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

$0.5=\frac{P_A}{16kPa}$

$P_A=8kPa$

Now we have to calculate the molar saturation.

The formula used for molar saturation is:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

and,

P(vapor free) = Total pressure - Vapor pressure

P(vapor) = $P_A$ = 8 kPa

So,

P(vapor free) = 100 kPa - 8 kPa = 92 kPa

The molar saturation will be:

$\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}$

$\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087$

Therefore, the molar saturation is 0.087