After adding a solute, freezing point of water decreases to –0.186 ⁰C. What is the
value of $\mathrm{\Delta }$T_b ? (k_b = 0.521; k_f = 1.86)

Given, k_b = 0.521; k_f = 1.86

${\mathrm{\Delta T}}_{\mathrm{f}}=0.186$

We know that, ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}$

Rearranging the formula we have:

$\mathrm{m}=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{0.186}{1.86}=0.1$

${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{k}}_{\mathrm{b}}\times \mathrm{m}=0.521\times 0.1=0.0521$

Hence, the required answer is B) 0.0521.