chemistry help!

Question

what is the boiling point of a solution of .75 mol of KBr in 1.0 kg of water?

Answer 1

This question demonstrates the colligative properties of a solution. Use the equation, ΔT_b = K_b x molality to solve this problem.

ΔT_b = T_b solution - T_b solvent

Solute = KBr

Solvent = H₂O

T_b of solvent = 100.00 °C

K_b for water = 0.512 °C m^-1

molality (m) = moles of solute / kg of solvent

m = 0.75 mol / 1.0 kg = 0.75 m

Enter the values from above in the equation,

ΔT_b = K_b x m

T_b solution - 100.00 = 0.512 x 0.75

T_b solution = 100.00 + 0.384 = 100.38 °C