Ratio of boiling points of aqueous solution containing 0.33 g of NaOH and another aqueous solution containing 3 g of glucose in the same volume is

For NaOH:

Given mass = 0.33 g 

Molar mass = 40 g

moles = $\frac{0.33}{40}$

For glucose (C₆H₁₂O₆)

Given mass = 3 g 

Molar mass = 180 g

moles = $\frac{3}{180}$

We can use:

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\mathrm{m}$

K_b and wt. of solvent is same in both cases so ratio of boiling points will be the ratio of the respective moles:

Ratio = $=\frac{\mathrm{moles} \mathrm{of} \mathrm{NaOH}}{\mathrm{moles} \mathrm{of} \mathrm{glucose}}=\frac{{\displaystyle \frac{0.33}{40}}}{{\displaystyle \frac{3}{180}}}=0.50$

Hence, the required answer is C) 0.50.