Verified Solution • 2025–26 Session moderate

QJEEM15Ch41d6: JEE Main 2015 Chemistry - Class 12 JEEM15Ch41d6 Solution: The vapour pressure of acetone at 20 o C is 185 torr. When 1.2…

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:

Multiple Choice Options

  • A 64
  • B 128
  • C 488
  • D 32

Step-by-Step Verified Solution

Using relation,

P∘−PsPs=w2M1w1M2{{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}

where w1,{w_1}, M1={M_1} = mass in gg

and mol. mass of solvent

w2,M2={w_2},{M_2} = mass in gg

and mol. mass of solute

Let M2=x{M_2} = x

P∘=185  torr;  Ps=183 torr{P^ \circ } = 185\,\,torr;\,\,{P_s} = 183\,torr

185−183183=1.2×58100x{{185 - 183} \over {183}} = {{1.2 \times 58} \over {100x}}

(Mol. mass of acetone =58=58 )

x=64x=64

∴\therefore      \,\,\,\,\, Molar mass of substance =64=64

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