JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 12)
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = $${3 \over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
$${4 \over 3}$$
$${3 \over 4}$$
$${1 \over 2}$$
$${1 \over 4}$$
Explanation
Relative lowering of vapour pressure,
$${{\Delta P} \over P}$$ = $${{{n_2}} \over {{n_1}}}$$
n2 = Number of moles of solute
n1 = Number of moles of solvent.
Given that,
Here is 5 molal solution, means 5 moles of solute are dissolved in 1 kg or 1000 g of solvent.
$$\therefore\,\,\,$$ Number of moles of solute = 5
Number of moles of solvent X = $${{1000} \over {{M_X}}}$$
Number of moles of solvent Y = $${{1000} \over {My}}$$
$$\therefore\,\,\,$$ $${\left( {{{\Delta \,P} \over P}} \right)_x}$$ = $${5 \over {{{1000} \over {{M_x}}}}}$$ = $${{5{M_x}} \over {1000}}$$
$${\left( {{{\Delta \,P} \over P}} \right)_y}$$ = $${5 \over {{{1000} \over {{M_y}}}}}$$ = $${{5\,{M_y}} \over {1000}}$$
Accoding to the question.
$${{5{M_x}} \over {1000}}$$ = m $$ \times $$ $${{5\,{M_y}} \over {1000}}$$
$$ \Rightarrow $$$$\,\,\,$$ Mx = m $$ \times $$ My
$$ \Rightarrow $$$$\,\,\,$$ $${3 \over 4}$$ My = m $$ \times $$ My [as given, Mx = $${3 \over 4}$$ My]
$$ \Rightarrow $$$$\,\,\,$$ m = $${3 \over 4}$$
$${{\Delta P} \over P}$$ = $${{{n_2}} \over {{n_1}}}$$
n2 = Number of moles of solute
n1 = Number of moles of solvent.
Given that,
Here is 5 molal solution, means 5 moles of solute are dissolved in 1 kg or 1000 g of solvent.
$$\therefore\,\,\,$$ Number of moles of solute = 5
Number of moles of solvent X = $${{1000} \over {{M_X}}}$$
Number of moles of solvent Y = $${{1000} \over {My}}$$
$$\therefore\,\,\,$$ $${\left( {{{\Delta \,P} \over P}} \right)_x}$$ = $${5 \over {{{1000} \over {{M_x}}}}}$$ = $${{5{M_x}} \over {1000}}$$
$${\left( {{{\Delta \,P} \over P}} \right)_y}$$ = $${5 \over {{{1000} \over {{M_y}}}}}$$ = $${{5\,{M_y}} \over {1000}}$$
Accoding to the question.
$${{5{M_x}} \over {1000}}$$ = m $$ \times $$ $${{5\,{M_y}} \over {1000}}$$
$$ \Rightarrow $$$$\,\,\,$$ Mx = m $$ \times $$ My
$$ \Rightarrow $$$$\,\,\,$$ $${3 \over 4}$$ My = m $$ \times $$ My [as given, Mx = $${3 \over 4}$$ My]
$$ \Rightarrow $$$$\,\,\,$$ m = $${3 \over 4}$$
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