What is relative lowering of vapour pressure derive the relationship between the relative lowering of vapour pressure and molecular mass solute?

Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g

Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2

= 60 g mol–1
According to Raoult's law, p0-pp0=ωMWm

p=p0-w×Mm×W×p°

p=23.8-50×1860×850

=23.8-0.017=23.78

Hence, 23.78 mm Hg. Ans.

17. Relation between relative lowering of vapour pressure and molality

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Let W2 g of solute of molar mass M2 be dissolved in W1 g of solvent of molar mass M1 . Hence number of
mole of solvent n1 and number of mole of solute n2 in solution.

`n_1=W_1/M_1 ` and `n_2=W_2/M_2` `(because "Number of moles (n)"="mass of the substance"/"molar mass of the substance")`

The mole fraction of solute x2 is given by

`x_2=n_2/(n_1+n_2)`

`x_2=(W_2/M_2)/(W_1/M_1+W_2/M_2)` .....(1)

For a solution of two components A1 and A2 with mole fraction x1 and x2 respectively, if the vapour pressure of pure component A1 is

`P_1^0 ` and that of component A2 is p_2^0 The relative lowering of vapour pressure is given by,

`(Deltap)/p_0^1=(p_1^0-p)/p_1^0`

`(Deltap)/p_0^1=(p_1^0x_2)/p_1^0`

`(Deltap)/p_0^1=x_2` .....(2)

Combining equations (1) and (2)

`(Deltap)/p_0^1=(p_1^0-p)/p_1^0=x_2=(W_2/M_2)/(W_1/M_1+W_2/M_2)`

For dilute solutions n1 >> n2. Hence n2 may be neglected in comparison with n1 in equation (1) and thus equation (3) becomes

`(Deltap)/p_0^1=n_2/n_1=(W_2/M_2)/(W_1/M_1)=(W_2M_1)/(W_1M_2)`

Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally vapour pressure of pure solvent and the solution, it is possible to determine molar mass of a non-volatile solute.

Answer

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Hint: Relative lowering of vapour pressure is the ratio of the vapour pressure lowering of the solvent from solution to the vapour pressure of the pure solvent. This lowering in vapour pressure has occurred when a solute is added to the solvent.

Complete step by step answer:

The expression for the relative lowering of vapour pressure can be written as;\[\dfrac{\Delta P}{{{P}^{{}^\circ }}}=\dfrac{{{P}^{{}^\circ }}-P}{{{P}^{{}^\circ }}}\]\[\Delta P = \]Lowering in vapour pressure\[{{P}^{{}^\circ }}=\]Vapour pressure of the pure solvent\[P= \]Vapour pressure of the solvent from the solutionAlso, we have known that the relative lowering of vapour pressure is equal to the mole fraction of the solute(\[{{x}_{2}}\]). Hence we can be written as;\[\Rightarrow \dfrac{\Delta P}{{{P}^{{}^\circ }}}=\dfrac{{{P}^{{}^\circ }}-P}{{{P}^{{}^\circ }}}={{x}_{2}}\]Hence, the mole fraction can be written as;\[\Rightarrow {{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{2}}+{{n}_{2}}}=\dfrac{\dfrac{{{W}_{2}}}{{{M}_{2}}}}{\dfrac{{{W}_{2}}}{{{M}_{2}}}+\dfrac{{{W}_{1}}}{{{M}_{1}}}}\]\[{{n}_{1}}\] and \[{{n}_{2}}\] are the number of moles of solute and solvent, respectively. \[{{W}_{1}}\] and \[{{M}_{1}}\] are the mass and molar mass of the solvent, respectively. \[{{W}_{2}}\] and \[{{M}_{2}}\] are the mass and molar mass of the solute, respectively. We have known that for dilute solutions, \[{{n}_{1}}>>{{n}_{2}}\]Hence, \[{{n}_{1}}\] is too small. So we can neglect the value of \[{{n}_{1}}\] when compared with \[{{n}_{2}}\]Hence , the equation for the mole fraction of solute becomes;\[\Rightarrow {{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{2}}+}=\dfrac{\dfrac{{{W}_{2}}}{{{M}_{2}}}}{\dfrac{{{W}_{1}}}{{{M}_{1}}}}\]So, finally we got the equation for the mole fraction of solute in terms of molar mass of solute and solvent respectively. Also, we have known that the mole fraction of solute is equal to the relative lowering of vapour pressure. So we can equate both equations;\[\Rightarrow \dfrac{\Delta P}{{{P}^{{}^\circ }}}=\dfrac{{{P}^{{}^\circ }}-P}{{{P}^{{}^\circ }}}=\dfrac{\dfrac{{{W}_{2}}}{{{M}_{2}}}}{\dfrac{{{W}_{1}}}{{{M}_{1}}}}\]Hence, it derived the relationship between the relative lowering of vapour pressure and molar mass of the non-volatile solute.

Note: Relative lowering of vapour pressure is colligative property (which depends upon the number of particles only. Other than this, there are another three types of colligative properties such as boiling point elevation, depression in freezing point, and osmotic pressure.