Question

derive a relation between relative lowering of Vapour pressure and molar mass of a solute.​

Answer 1

Here, P0ΔP and P0P0−P represents relative lowering of vapour pressure and X2 represents mole fraction of solute. Here, W1 and M1 are the mass and molar masses of solvent and W2 and M2 are the mass and molar masses of solute. n1 and n2 are the number of moles of solute and solvent respectively.

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Answer 2

$\huge\star\mathfrak\pink{{Answer:-}}$

Consider a solution in which W2

g of solute of molar mass M2 is dissolved in W1

g of solvent of molar mass M1

.

Hence, number of moles of solvent, n1=M1 W1

and,

Number of moles of solute, n1=M2W2

∴ Total number of moles n=n1+n2

Mole fraction of the solvent, xx1 =nn1

Mole fraction of the solute,x x2 =nn 2

In case of a dilute solution, the concentration of number of moles of the solute is very low, i.e., n1>>n2

∴n1 +nn ≃n1

Now, the mole fraction of the solute x2

is given by,

x2 =n1 +n2

n2≃ n1 n2

∴x2 = W1 /M1

W2 /M2

∴x2 =W1 ×M2

W2×M1

lf P1

0

and P are the vapour pressure of pure solvent and a solution respectively, then relative lowering of vapour pressure is given by,

P= P1 0

P1 0−P

By Raoult's law

P1 0

ΔP =P1 0

P1 0 −P =x2

P 1 0

P1 0−P= W1 M2 W2M11

or ∴

P

1

0

ΔP

=

n

1

n

2

=

W

1

/M

1

W

2

/M

2

=

W

1

M

2

W

2

M

1

Hence, by determining the vapour pressure of a pure solvent and a solution, the molar mass of a non-volatile solute can be measured.

Hope It helps...

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