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Question: The mass of glucose that should be dissolved in \( 100 \) g of water in order to produce the same lo...

The mass of glucose that should be dissolved in 100100 g of water in order to produce the same lowering of vapour pressure as produced by dissolving 1g1g urea in 50g50g of water is?
A. 1g1g
B. 2g2g
C. 6g6g
D. 12g12g

Explanation

Solution

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 =W2M1W1M2= \dfrac{{{W_2}{M_{_1}}}}{{{W_1}{M_2}}}
W1{W_1} = Mass of solute
W2{W_2} = Mass of solvent
M1{M_{_1}} = Molar mass of solute
M2{M_2} = Molar mass of solvent.

Complete Step By Step Answer:
Given:
For Glucose solution
Mass of water ( W2{W_2} ) =100g= 100g
For Urea solution
Mass of Urea ( W1{W_1} ) =1g= 1g
Mass of water ( W2{W_2} ) =50g= 50g
To find = Mass of glucose ( W1{W_1} )
Molecular mass of glucose =180g= 180g
Molecular mass of urea =60g= 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 =W2M1W1M2= \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
(W2M1W1M2)glucose=(W2M1W1M2)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,
W1×18100×180=1×1850×60\dfrac{{{W_1} \times 18}}{{100 \times 180}} = \dfrac{{1 \times 18}}{{50 \times 60}}
W1=100×18050×60{W_1} = \dfrac{{100 \times 180}}{{50 \times 60}}
On solving the above equation, we get,
W1=6g\Rightarrow {W_1} = 6g
Therefore, the mass of glucose that should be dissolved in 100100 g of water is 6g6g .
Hence, the correct option is C. 6g6g .

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.