JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Some Basic Concepts of Chemistry

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 density of the solution in

gmL^{- 1}, C_{2}

is the molarity and

x_{2}

is the mole fraction of the

2^{nd}

component, then

C_{2}

can be expressed as:

(A) $C_{2} = \frac{1000 x _{2}}{M _{1} + x _{2} ( M _{2} - M _{1} )}$
(B) $C_{2} = \frac{dx _{2}}{M _{2} + x _{2} ( M _{2} - M _{1} )}$
(C) $C_{2} = \frac{1000 dx _{2}}{M _{1} + x _{2} ( M _{2} - M _{1} )}$ ✓ Correct
(D) $C_{2} = \frac{dx _{1}}{M _{2} + x _{2} ( M _{2} - M _{1} )}$

Answer: (C)

Solution

Work with one total mole of solution, of which

x_{2}

is the second component. Its mass is

x_{2} M_{2} + (1 - x_{2}) M_{1} = M_{1} + x_{2} (M_{2} - M_{1})

, and dividing by the density

d

(in

g mL^{- 1}

) gives the volume

\frac{M _{1} + x _{2} ( M _{2} - M _{1} )}{d} mL

, i.e.

\frac{M _{1} + x _{2} ( M _{2} - M _{1} )}{1000 d} L

. Molarity is moles of the component per litre, so

C_{2} = \frac{x _{2}}{\frac{M _{1} + x _{2} ( M _{2} - M _{1} )}{1000 d}} = \frac{1000 d x _{2}}{M _{1} + x _{2} ( M _{2} - M _{1} )}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option