A solution of two components containing n _ 1 moles of the 1 ^ st component and n _ 2 moles of the 2 ^ nd component is prepared. M _ 1 and M _ 2 are the molecular weights of component 1 and 2 respectively. If d is the

Correct answer: C. C _ 2 = 1000 dx _ 2 M _ 1 + x _ 2 ( M _ 2 - M _ 1 ) C _ 2 = 1000 dx _ 2 M _ 1 + x _ 2 ( M _ 2 - M _ 1 )

JEE Main 20206 SeptemberMorning Shift

JEE Main 2020 · 6 September · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

A solution of two components containing n _ 1 moles of the 1 ^ st...

A solution of two components containing

{n}_{1}

moles of the

{1}^{\text{st}}

component and

{n}_{2}

moles of the

{2}^{\text{nd}}

component is prepared.

{M}_{1}

and

{M}_{2}

are the molecular weights of component 1 and 2 respectively. If

d

is the density of the solution in

{gmI}^{- 1}, {C}_{2}

is the molarity and

{x}_{2}

is the mole fraction of the

{2}^{\text{nd}}

component, then

{C}_{2}

can be expressed as :

A{C}_{2} = \dfrac{1000{x}_{2}}{{M}_{1} + {x}_{2} $\left({M}_{2} - {M}_{1}\right)}$
B{C}_{2} = \dfrac{{dx}_{2}}{{M}_{1} + {x}_{2} $\left({M}_{2} - {M}_{1}\right)}$
C{C}_{2} = \dfrac{1000{dx}_{2}}{{M}_{1} + {x}_{2} $\left({M}_{2} - {M}_{1}\right)}$
D{C}_{2} = \dfrac{{dx}_{1}}{{M}_{2} + {x}_{2} $\left({M}_{2} - {M}_{1}\right)}$

C. {C}_{2} = \dfrac{1000{dx}_{2}}{{M}_{1} + {x}_{2} $\left({M}_{2} - {M}_{1}\right)}$

{C}_{2} = \dfrac{1000{dx}_{2}}{{M}_{1} + {x}_{2}

\left({M}_{2} - {M}_{1}\right)}