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At room temperature, a dilute solution of urea is prepared by dissolving 0.60g of urea in 360g of water. If the vapour pressure of pure water at this temperature is 35mmHg. Lowering of vapour pressure will be (molar mass of urea = $60 g mol^{-1}$)
(a). 0.02mmHg
(b). 0.028mmHg
(c). 0.017mmHg
(d). 0.031mmHg
Answer
497.4k+ views
Hint: The above numerical is based on one of colligative properties of the solution i.e. relative lowering of vapour pressure. The lowering in vapour pressure is due to the fact that after the solute was added to the pure liquid (solvent), the liquid surface now had molecules of both, the pure liquid and the solute.
Complete step by step solution:
There are certain properties of dilute solutions which depend only on the number of particles of solute and not on the nature of the solvent and the solute. Such properties are called colligative properties. There are four colligative properties: Relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.
We will now discuss relative lowering of vapour pressure in the following section in detail.
According to Raoult’s law for solutions containing non-volatile solute
When a non-volatile solute in a pure solvent is dissolved, the vapour pressure of a pure solvent gets gradually decreased. Mathematically, it can be represented as-
\[\dfrac{{{\text{P}}_{\text{A}}}^{\text{o}}-{{\text{P}}_{\text{A}}}}{{{\text{P}}_{\text{A}}}^{\text{o}}}={{\text{X}}_{\text{B}}}\] , where, ${{\text{X}}_{\text{B}}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}$
or, $\Delta\text{P=}{{\text{P}}^{\text{o}}}\text{x }\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}$
$=35\text{x}\dfrac{\dfrac{0.6}{60}}{\dfrac{360}{18}+\dfrac{0.6}{60}}=\dfrac{35\text{x}100}{2001\text{x100}}=0.017\text{mmHg}$
Therefore, the correct option is (c) 0.017mmHg as per the calculation.
Note: As we have seen, the above expression can be used to determine the molecular mass of the solute B, provided the relative lowering of vapour pressure of a solution of known concentration and molecular mass of the solvent are known. However, the determination of molecular mass by this method is often difficult because the accurate determination of lowering of vapour pressure is difficult.
Complete step by step solution:
There are certain properties of dilute solutions which depend only on the number of particles of solute and not on the nature of the solvent and the solute. Such properties are called colligative properties. There are four colligative properties: Relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure.
We will now discuss relative lowering of vapour pressure in the following section in detail.
According to Raoult’s law for solutions containing non-volatile solute
When a non-volatile solute in a pure solvent is dissolved, the vapour pressure of a pure solvent gets gradually decreased. Mathematically, it can be represented as-
\[\dfrac{{{\text{P}}_{\text{A}}}^{\text{o}}-{{\text{P}}_{\text{A}}}}{{{\text{P}}_{\text{A}}}^{\text{o}}}={{\text{X}}_{\text{B}}}\] , where, ${{\text{X}}_{\text{B}}}=\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}$
or, $\Delta\text{P=}{{\text{P}}^{\text{o}}}\text{x }\dfrac{{{\text{n}}_{\text{B}}}}{{{\text{n}}_{\text{A}}}+{{\text{n}}_{\text{B}}}}$
$=35\text{x}\dfrac{\dfrac{0.6}{60}}{\dfrac{360}{18}+\dfrac{0.6}{60}}=\dfrac{35\text{x}100}{2001\text{x100}}=0.017\text{mmHg}$
Therefore, the correct option is (c) 0.017mmHg as per the calculation.
Note: As we have seen, the above expression can be used to determine the molecular mass of the solute B, provided the relative lowering of vapour pressure of a solution of known concentration and molecular mass of the solvent are known. However, the determination of molecular mass by this method is often difficult because the accurate determination of lowering of vapour pressure is difficult.
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