Question

30g of urea is added to 45g of H2O at a certain temperature. If vapour pressure of pure water at that pressure is 300mm of Hg. Then calculate-
1) Mole fraction of Xa and Xb
2) Vapour pressure of solution
3) Lowering in Vapour pressure
4) RLVP
5) Molality

Answer 1

Answer:

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into a foil

2. Property of a substance which can be drawn

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Explanation:

1. Property of a substance which can be beaten

into a foil

2. Property of a substance which can be drawn

into wire1. Property of a substance which can be beaten

into a foil

2. Property of a substance which can be drawn

into wire

Answer 2

The Mole fraction of Xa and Xb, Vapour pressure of solution are respectively 0.1667, 0.8333 and 50.01pa.

Given:-

Mass of urea = 30g

Mass of water = 45g

Pressure = 300mm of Hg

To Find:-

Mole fraction of Xa and Xb, Vapour pressure of solution.

Solution:-

We can easily find out the Mole fraction of Xa and Xb, Vapour pressure of solution, Lowering in Vapour pressure, RLVP and Molality of the solution by following these simple steps.

Now,

Mass of urea = 30g

Mass of water = 45g

Pressure = 300mm of Hg

Mole Fraction of xa and xb,

For finding the mole fraction, first we have to find out the number of moles of urea and water.

Moles of urea =

$\frac{given \: mass}{molar \: mass}$

$n = \frac{m}{mm}$

$n = \frac{30}{60}$

$n = \frac{1}{2}$

n(urea) = 0.5

Moles of water =

$\frac{mass \: of \: water}{molar \: mass}$

$n = \frac{45}{18}$

$n = \frac{5}{2}$

n(water) = 2.5

Mole Fraction of urea (xa) =

$\frac{moles \: of \: urea}{total \: moles}$

$xa = \frac{0.5}{0.5 + 2.5}$

$xa = \frac{0.5}{3.0}$

xa = 0.1667

Mole Fraction of Water (xb) = 1 - xa

$xb = 1 - xa$

$xb = 1 - 0.1667$

xb = 0.8333

Vapour pressure of the Solution,

Vapour pressure of solution = mole fraction of solvent* Vapour pressure of Pure water

$p = xa \times po$

$p = 0.1667 \times 300$

p = 50.01 pa

Hence, the Mole fraction of Xa and Xb, Vapour pressure of solution are respectively 0.1667, 0.8333 and 50.01pa.

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