Freezing point depression to find molecular weight
Mass of solute $W_b=0.845\times10^{-3}\;kgm$
  Mass of solvent $W_a=4.250\times10^{-3}\;kgm$
  Depression in freezing point $\Delta T_f=1.36\;K$
  Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$
  Freezing point of Camphor $T_f=129\;^\circ C$




Part I: $\Delta T_f$

Part II: Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$

Part III: Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$

Part IV: Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$