Question

5)The freezing point of a solution made by dissolving 14.og
unknown in 200g of water (Kf = 1.86°C/m, Tf = 0°C)
decreased by 3°C, the molar mass of unknown is :​

Answer 1

The change in freezing point is given by,

$\tt{\longrightarrow\Delta_fT=ik_fm}$

$\tt{\longrightarrow 3=1\times1.86m}$

$\tt{\longrightarrow m=\dfrac{3}{1.86} }$

That is,

$\tt{\longrightarrow\dfrac{1000W_B}{M_B\cdot W_A}=\dfrac{3}{1.86}}$

Here,

• $\tt{W_B=14\ g}$

• $\tt{W_A=200\ g}$

Then,

$\tt{\longrightarrow\dfrac{1000\times14}{200M_B}=\dfrac{3}{1.86} }$

$\tt{\longrightarrow M_B=\dfrac{70\times1.86}{3}}$

$\tt{\longrightarrow\underline{\underline{M_B=43.4\ g\,mol^{-1}}}}$