Question

Derive relationship between molarity and mole fraction

Answer 1

Answer:

hey mate

Explanation:

here is your answer

Relationship between molarity and mole fraction

$m = \frac{n \times 100}{v \: in \: ml}$

$d = \frac{m}{v}$

M = mass of solution

V = volume of solution

mass of solution = na 1 ma 1 + NB + MB

$d \: = \frac{na \: ma \: + \: nb \: mb \: }{v}$

$v \: = \frac{na \: ma \: + nb \: mb \: }{d}$

$m = \frac{na \: \times 1000}{ \frac{na \: mn \: + \: nb \: mb \: }{d} }$

$m = \frac{na \times 1000}{na \: ma \: + \: nb \: mb}$

In place of both cases

Na and N b

divide 1 term with Na + NB

$m \: = \: \frac{n \: \times \: 1000 \times d}{na + nb}$

$\frac{na}{na \: + \: nb} \times ma \: + \frac{nb}{na \: + nb} \times mb$

we know that

$m \: = \frac{ \infty a \: \times \: 1000 \times d}{xa \: \times ma \times xb \: \times mb}$

$m \: = \frac{xa \: \times 1000 \: \times d}{xa \: ma \: + xb \: mb}$

XA = mole fraction of solute

XB = mole fraction of solvent

$thanku$