Get Answers to all your Questions

header-bg qa

Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
(a) CHCl_3(l) \; \text{and} \;CH_2Cl_2(l)
(b) NaCl(s)\; \text{and}\;H_{2}O(l)

Answers (1)

p=p_A^ox_A+p_B^ox_B

Where, p_A^o,p_B^o are the vapour pressure of pure components (A) and (B)

And x_{A}\text{and }x_{B} are the mole fractions of the components in the solution.

 

(b) On dissolving in water, the vapour pressure lowers. Vapour pressure of solution is:

p=p^ox_A

Where x_{A}=mole fraction of solvent

p^{0}=vapour pressure of pure solvent

p=vapour pressure of solution

Similarly,

\frac{\Delta p}{p^o}=x_B

\Delta p=p^o-p

x_{B}=mole fraction of solute

Posted by

infoexpert24

View full answer