(યુરિયા અને ગ્લુકોઝનો અણુભાર અનુક્રમે $60$ અને $180$ છે.)
$10\, g$ of urea; $m =\frac{\frac{1}{60}}{ v }=\frac{1}{60 v }$
$3\, g$ of glucose; $m =\frac{\frac{3}{180}}{ v }=\frac{1}{60 v }$
We know
$\Delta T _{ b }= k _{ b } \times \,m$
$\Rightarrow \Delta T _{ b } \propto m$
Here, molarity remains unchanged
Hence, they have the same boiling point.
Hence, the correct option is $c$.
[આપેલ $K_b (H_2O) = 0.52\, K\, kg\, mol^{-1}]$