Question

If glucose of $36\, g$ weight is dissolved in $2 \,kg$ of $H_2O$ then, change in boiling point $(\Delta T_b)$ at 1.013 bar will be ($K_b$ for ${H_2O}$ is ${ 0.52\, K\, kg \,mol^{-1}}$)

• A. 1.04 K
• B. 0.52 K
• C. 0.052 K
• D. 5.2 K

Answer 1

Answer: (C)

0.052 K

Answer 2

$\Delta T_b = K_b \times m$
$\Delta T_b = 0.52 \times \left( \frac{36 \, g}{180 \times 2}\right)= 0.052\,{ K}$