The expression relating to mole fraction of solute $(x_{2})$ and molarity (M) of the solution is:

x_{2} = \frac{M M_{1}}{M (M_{1} - M_{2}) + ρ}

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x_{2} = \frac{M M_{1}}{M (M_{1} - M_{2}) - ρ}

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x_{2} = \frac{M (M_{1} - M_{2}) + ρ}{M M_{1}}

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x_{2} = \frac{M (M_{1} - M_{2}) - ρ}{M M_{1}}

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Solution

The correct option is A $x_{2} = \frac{M M_{1}}{M (M_{1} - M_{2}) + ρ}$
Let us consider the problem.
$x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = \frac{n_{2}}{(m_{1} / M_{1}) + n_{2}} = \frac{n_{2} M_{1}}{m_{1} + n_{2} M_{1}}$
Hence,
$= \frac{n_{2} M_{1}}{(m_{1} + m_{2}) + n_{2} M_{1} - m_{2}}$
Now, as $ρ = \frac{m a s s}{v o l u m e}$ and $M = n m$
$∴ x_{2} = \frac{n_{2} M_{1}}{V ρ + n_{2} (M_{1} - M_{2})} = \frac{(n_{2} / V) M_{1}}{ρ + (n_{2} / V) (M_{1} - M_{2})}$
$= \frac{M M_{1}}{ρ + M (M_{1} - M_{2})}$

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The expression relating to mole fraction of solute (x2) and molarity (M) of the solution is:

The correct option is A x2=MM1M(M1−M2)+ρLet us consider the problem.x2=n2n1+n2=n2(m1/M1)+n2=n2M1m1+n2M1Hence,=n2M1(m1+m2)+n2M1−m2Now, as ρ=massvolume and M=nm∴x2=n2M1Vρ+n2(M1−M2)=(n2/V)M1ρ+(n2/V)(M1−M2)=MM1ρ+M(M1−M2)

The expression relating to mole fraction of solute $(x_{2})$ and molarity (M) of the solution is: