Verified Solution • 2025–26 Session moderate

QJEEM18Ch9828: JEE Main 2018 Chemistry - Class 12 JEEM18Ch9828 Solution: Two 5 molal solutions are prepared by dissolving a…

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 = 34{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 :

Multiple Choice Options

  • A 43{4 \over 3}
  • B 34{3 \over 4}
  • C 12{1 \over 2}
  • D 14{1 \over 4}

Step-by-Step Verified Solution

Relative lowering of vapour pressure,

    ΔPP{{\Delta P} \over P}   =   n2n1{{{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  =   1000MX{{1000} \over {{M_X}}}

Number of moles of solvent Y = 1000My{{1000} \over {My}}

∴   \therefore\,\,\, (Δ PP)x{\left( {{{\Delta \,P} \over P}} \right)_x}   =   51000Mx{5 \over {{{1000} \over {{M_x}}}}} = 5Mx1000{{5{M_x}} \over {1000}}

(Δ PP)y{\left( {{{\Delta \,P} \over P}} \right)_y}   =   51000My{5 \over {{{1000} \over {{M_y}}}}}   =   5 My1000{{5\,{M_y}} \over {1000}}

Accoding to the question.

5Mx1000{{5{M_x}} \over {1000}}   =   m ×\times 5 My1000{{5\,{M_y}} \over {1000}}

⇒\Rightarrow    \,\,\, Mx  =   m ×\times My

⇒\Rightarrow    \,\,\, 34{3 \over 4} My = m ×\times My   [as given,   Mx = 34{3 \over 4} My]

⇒\Rightarrow    \,\,\, m   =   34{3 \over 4}

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