Answer
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Hint: When a solute is added to the solvent the vapour pressure of the solvent decreases. This is known as the lowering of vapour pressure. The lowering of the vapour pressure depends upon the amount of non-volatile solute is added in the solution.
Complete step by step answer: It is given is the question that two solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y.
It is also given that $ { M }_{ X } = \cfrac { 3 }{ 4 } { M }_{ Y }$ ------ (1)
We know that the relative lowering of the vapour pressure of two solutions is given as:
$ { (\cfrac { \Delta P }{ P } ) }_{ X } = m{ (\cfrac { \Delta P }{ P } ) }_{ Y }$
The above relation can be written because it is given that the relative lowering of the vapour pressure of the solution X in "m" times that of the solution in Y.
Now, we know that the relative lowering of the vapour pressure of a solution is directly proportional to the mole fraction of the solution. Thus, from the vapour pressure relation, we get
$ { M }_{ x } \times \cfrac { 5 }{ 1000 } = m \times { M }_{ Y } \times \cfrac { 5 }{ 1000 }$
Here, $\cfrac {5}{1000}$ gets cancelled from both the sides. And then substituting this equation in equation (1), we get
$ \cfrac { 3 }{ 4 } \times { M }_{ Y } = m \times { M }_{ Y }$
Here, ${M}_{Y}$ gets cancelled from both the sides.
$ m = \cfrac { 3 }{ 4 }$
Therefore, the value of m is $\cfrac {3}{4}$. Hence, the correct answer is option (A).
Note: 5 molal solutions of both X and Y are prepared. It means that 5 moles of the solute is dissolved in 1 kg (or 1000g) of the solvent.
Therefore, no. of moles of solvent becomes $\cfrac { 1000 }{ M }$
Thus, the mole fraction of the solvent becomes $\cfrac { 5 }{ \cfrac { 1000 }{ M } }$
Complete step by step answer: It is given is the question that two solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y.
It is also given that $ { M }_{ X } = \cfrac { 3 }{ 4 } { M }_{ Y }$ ------ (1)
We know that the relative lowering of the vapour pressure of two solutions is given as:
$ { (\cfrac { \Delta P }{ P } ) }_{ X } = m{ (\cfrac { \Delta P }{ P } ) }_{ Y }$
The above relation can be written because it is given that the relative lowering of the vapour pressure of the solution X in "m" times that of the solution in Y.
Now, we know that the relative lowering of the vapour pressure of a solution is directly proportional to the mole fraction of the solution. Thus, from the vapour pressure relation, we get
$ { M }_{ x } \times \cfrac { 5 }{ 1000 } = m \times { M }_{ Y } \times \cfrac { 5 }{ 1000 }$
Here, $\cfrac {5}{1000}$ gets cancelled from both the sides. And then substituting this equation in equation (1), we get
$ \cfrac { 3 }{ 4 } \times { M }_{ Y } = m \times { M }_{ Y }$
Here, ${M}_{Y}$ gets cancelled from both the sides.
$ m = \cfrac { 3 }{ 4 }$
Therefore, the value of m is $\cfrac {3}{4}$. Hence, the correct answer is option (A).
Note: 5 molal solutions of both X and Y are prepared. It means that 5 moles of the solute is dissolved in 1 kg (or 1000g) of the solvent.
Therefore, no. of moles of solvent becomes $\cfrac { 1000 }{ M }$
Thus, the mole fraction of the solvent becomes $\cfrac { 5 }{ \cfrac { 1000 }{ M } }$
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