Question

the vapour pressure of benzene is 200nm .the lowering of vapour pressure of 0.2 molal solution is (in mm)​

Answer 1

Explanation:

mole fraction of solute can be calculated by

x2 =mxm / 1000

x2 = 0.2x78/1000

x2=15.6x10 -3

x2=0.0156

relative lowering of vapour pressure is calculated by

p/p =x2

p/p = 0.0156

hence relative lowering of vapour pressure is 0.0156

Answer 2

Answer:

The lowering of the vapour pressure, Δp, measured is $3.12 \times 10^{-6} mm$. 

Explanation:

Given,

The vapour pressure of benzene, $p\textdegree$ = $200nm$ = $200 \times 10^{-6}mm$ = $2\times 10^{-4}mm$

The molality of the solution, m = $0.2molal$

The lowering of the vapour pressure, Δp =?

Now,

• From the equation given below, we can find out the lowering of the vapour pressure:
• $\frac{\Delta p}{p\texrdegree} = \frac{m\times M}{1000}$

Here,

• Δp = Lowering of the vapour pressure
• $p\textdegree$ = Vapour pressure
• m = Molality
• M = The molar mass of benzene = $78 g/mol$

After putting the given values in the equation, we get:

• $\frac{\Delta p}{2\times 10^{-4} } = \frac{0.2\times 78}{1000}$
• Δp = $31.2 \times 10^{-7} mm$ or $3.12 \times 10^{-6} mm$

Hence, the lowering of the vapour pressure, Δp = $3.12 \times 10^{-6} mm$.