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Hint: The pressure exerted by vapours over a liquid under equilibrium conditions at a certain temperature is known as vapour pressure. A solvent's vapour pressure in a solution is always lower than the pure solvent's vapour pressure. The decrease in vapour pressure is proportional to the mole fraction of the solute.
Relative lowering of vapour pressure $ = \dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}} $
$ {W_1} $ = Mass of solute
$ {W_2} $ = Mass of solvent
$ {M_{_1}} $ = Molar mass of solute
$ {M_2} $ = Molar mass of solvent.
Complete Step By Step Answer:
Given:
For Glucose solution
Mass of water ( $ {W_2} $ ) $ = 100g $
For Urea solution
Mass of Urea ( $ {W_1} $ ) $ = 1g $
Mass of water ( $ {W_2} $ ) $ = 50g $
To find = Mass of glucose ( $ {W_1} $ )
Molecular mass of glucose $ = 180g $
Molecular mass of urea $ = 60g $
From the question, we can say that
Mole fraction of solute in glucose = Mole fraction of solute in urea
Hence, vapour pressure lowering is same in both the cases
Using the above formula of Vapour pressure,
Relative lowering of vapour pressure $ = \dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}} $
We are given that
Relative lowering of vapour pressure of glucose = Relative lowering of vapour pressure of urea
$ {\left( {\dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}}} \right)_{glu\cos e}} = {\left( {\dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}}} \right)_{urea}} $
Substituting the given values in the above formula,
$ \dfrac{{{W_1} \times 18}}{{100 \times 180}} = \dfrac{{1 \times 18}}{{50 \times 60}} $
$ {W_1} = \dfrac{{100 \times 180}}{{50 \times 60}} $
On solving the above equation, we get,
$ \Rightarrow {W_1} = 6g $
Therefore, the mass of glucose that should be dissolved in $ 100 $ g of water is $ 6g $ .
Hence, the correct option is C. $ 6g $ .
Note:
Colligative properties are the characteristics of dilute solutions of non-volatile solutes that are determined by the concentration of solute particles in the solution rather than the chemical composition of the solute. Relative lowering of vapour pressure is a colligative property. Other colligative properties are Elevation of boiling point, Depression in freezing point and osmotic pressure.
Relative lowering of vapour pressure $ = \dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}} $
$ {W_1} $ = Mass of solute
$ {W_2} $ = Mass of solvent
$ {M_{_1}} $ = Molar mass of solute
$ {M_2} $ = Molar mass of solvent.
Complete Step By Step Answer:
Given:
For Glucose solution
Mass of water ( $ {W_2} $ ) $ = 100g $
For Urea solution
Mass of Urea ( $ {W_1} $ ) $ = 1g $
Mass of water ( $ {W_2} $ ) $ = 50g $
To find = Mass of glucose ( $ {W_1} $ )
Molecular mass of glucose $ = 180g $
Molecular mass of urea $ = 60g $
From the question, we can say that
Mole fraction of solute in glucose = Mole fraction of solute in urea
Hence, vapour pressure lowering is same in both the cases
Using the above formula of Vapour pressure,
Relative lowering of vapour pressure $ = \dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}} $
We are given that
Relative lowering of vapour pressure of glucose = Relative lowering of vapour pressure of urea
$ {\left( {\dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}}} \right)_{glu\cos e}} = {\left( {\dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}}} \right)_{urea}} $
Substituting the given values in the above formula,
$ \dfrac{{{W_1} \times 18}}{{100 \times 180}} = \dfrac{{1 \times 18}}{{50 \times 60}} $
$ {W_1} = \dfrac{{100 \times 180}}{{50 \times 60}} $
On solving the above equation, we get,
$ \Rightarrow {W_1} = 6g $
Therefore, the mass of glucose that should be dissolved in $ 100 $ g of water is $ 6g $ .
Hence, the correct option is C. $ 6g $ .
Note:
Colligative properties are the characteristics of dilute solutions of non-volatile solutes that are determined by the concentration of solute particles in the solution rather than the chemical composition of the solute. Relative lowering of vapour pressure is a colligative property. Other colligative properties are Elevation of boiling point, Depression in freezing point and osmotic pressure.
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