Question

81 Which solution has the higher boiling point, 0.500 M glucose or 0.300 M KCl? Explain.

Answer 1

Boiling point elevation(∆T_b) is calculated by the following formula

∆T_b = K_b × b × i

where,

K_b = boiling point elevation constant

b = molality of solute

i = Van't Hoff factor

For same solvent ∆T_b is proportianal to product of b×i

Molarity of glucose( 0.500M) is higher than molarity of KCl (0.300M), so molality of glucose is higher than molality of KCl. Molality of glucose is 1.6667 fold higher than molality of KCl but van't Hoff factor of KCl (2) is two fold higher than van't Hoff factor of glucose(1). So, b×i factor for KCl is higher than for glucose. So,

0.300M KCl solution has the higher boiling point.