When 3.00gm of a substance ‘x’ is dissolved in 100g of $CC{l}_4$  , it raises the boiling point by 0.60k the molar mass of the substance ‘x’ is _________ $gmo{l}^{-1}$(Given  $k_b$ for $CC{l}_4$  is 5.0 k kg $mo{l}^{-1}$ ) 
 

Let molar mass of “x” be M
           $△$Tb= kbM
                    = 5 X $\frac{WtX1000}{MXMassofSolvent\left(g\right)}$ 
                    $=5X\frac{3X1000}{MX100}$    $0.6=\frac{150}{M}$$M=\frac{150}{0.6}=250g/mole$