Question

The mole fraction of a solute in a solution is 0.1 at 298 kelvin molality of the solution is same as this molality density of the solution is 298 kelvin is 2 gram per centimetre cube the ratio of molecular weight of the solute and solvent is
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Answer 1

Question from JEE ADVANCED 2016 .

GiveN

• Mole fraction of solute in a solution is 0.1 .

• Molality of the solution is same as molarity.

• Density of the solution is 2g/cm³.

To Find

The ratio of molecular weight of the solute and solvent

Formulas involveD

Density = $\frac{mass }{volume}$

Number of moles = $\frac{Given\:weight }{Molar\:mass}$

Molarity = $\frac{Number\:of\:moles\:of\:solute }{Volume\:of\:solution\:in\:litre}$

Molality = $\frac{Number\:of\:moles\:of\:solute }{Weight\:of\:solvent\:in\:kg}$

CalculationS

• Let number of moles of solute be 0.1 x. Thus, number of moles of solvent will be ( 1 - 0.1)x = 0.9 x.

• Let the volume of the solution be V.

Density of the solution = 2 g/cm³ = 2 kg/m³

weight of the solution = density × volume = 2V.

Molarity = $\frac{ 0.1x }{V}$

Weight(W) of solvent = Number of moles × $M_{solvent}$

= 0.9 x $M_{solvent}$

Molality = $\frac{ 0.1x }{ 0.9 x M_{solvent} }$

Molarity = Molality ( Given )

=> $\frac{ 0.1x }{V} = \frac{ 0.1x }{ 0.9 x M_{solvent} }$

=> V = 0.9 x $M_{solvent}$

Now,

$W_{solute} + W_{solvent} = W_{solution}$

=> $0.1x M_{Solute} + 0.9 xM_{solvent} = 2V$

=> $0.1xM_{Solute} + 0.9 xM_{solvent} = 2(0.9M_{solvent} )$

=> $0.1xM_{Solute} + 0.9 xM_{solvent} = 1.8xM_{solvent}$

=> $0.1 M_{Solute} + 0.9M_{solvent} = 1.8M_{solvent}$

=> $0.1 M_{Solute} = 0.9M_{solvent}$

=> $\frac{0.9M_{Solvent}}{0.1M_{Solute}}$ = 1.

=> $\frac{M_{Solute}}{M_{Solvent}}$ = 9.

$\boxed{ \boxed{ \pink { \frac{M_{Solute}}{M_{Solvent}} = 9 } } }$