Question

calculate the vapour pressure of 0.1 molar urea solution at 298 K. the vapour pressure of pure water at 298 K is 20 mm of HG. assume density of solution to be 1 gcm-3​

Answer 1

Answer: The vapor pressure of solution is 19.96 mmHg.

Solution:-

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 20mmHg

$p_s$ = vapor pressure of solution  = ?

Given : Molarity of solution = 0.1 M = 0.1 moles of urea in 1000 ml of solution

Density of solution = $1g/cm^3$

Mass of solution=$Density\times Volume=1g/ml\times 1000ml=1000 g$

$w_2$ = mass of solute = $moles\times {\text {Molar mass}}=0.1\times 60g/mol=6 gram$

$w_1$ = mass of solvent  (water) = Mass of solution - mass of solute = 1000 - 6 = 994 gram

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (urea) = 60 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{20-p_s}{20}=\frac{6\times 18}{994\times 60}$

$p_s=19.96mm Hg$

Therefore, the vapor pressure of solution is, 19.96 mmHg.

Answer 2

Answer:

Vapour pressure of urea solution = 19.96 mmHg

Explanation:

Molarity = no. of moles of solute / 1L of solution

• 0.1 = no. of moles of solute/ 1L of solutiol

no. of moles = mass of solute/ molar mass of solute = 0.1

molar mass of solute (urea) = 60

mass of solute = 0.1 × 60 = 6 gram

• Mass of solution = 1000g

mass of solute + mass of solvent = 1000g

mass of solvent = 1000 - 6 = 994g

molar mass of solvent ( water) = 18

By using relative lowering in vapour pressure.

P°A - Ps / P°A = Wb/Mb × Ma/Wa

20 - Ps /20 = 6/60 × 18/ 994

Ps = 19.96mmHg