Discussion Forum

1)

 The freezing point  of solution containing  10 mL  of non-volatile and non-electrolyte liquid " A" in 500 g of water is $-0.413^{0} C$. If $K_{f}$ of water is 1.86 K kg $mol^{-1}$ and the molecular weight of A=60 g $mol^{-1}$, what is the density  of the  solution in g $mL^{-1}$ ?

 (Assume  $\triangle  _{mix}$v=0)

Answer:

Option D

Explanation:

Given,

 Volume of solute =10 mL

 $\triangle T_{f}=0-(-0.413)=0.413^{0}C$

 w(solvent) =500g

$K_{f}(water)$=1.86 K kg $mol ^{-1}$ ($=w_{A}$)

Molecular weight of (A) or $M_{B}$  = 60g

 Let mass of solute (A) = w_B

 $\therefore$    $\triangle T_{f}=K_{f}\times\frac{w_{B}}{M_{B}}\times\frac{1000}{w_{A}}$

        $0.413=1.86\times\frac{w_{B}}{60}\times\frac{1000}{500}$

 $\therefore$     $w_{B}=\frac{0.413 \times60\times500}{1.86\times 1000}$

 Mass of solute =6.66 g

 Also, $\because$   Density (d) of solution  = Total mass/ Total volume

 = $\frac{6.66+500}{500+10}= \frac{506.66}{510}$= 0.993 (g $mL^{-1}$)

 Density (d) of solution =0.993 g m L^-1

 Hence , option (d)  is the correct answer