A certain liquid X has a normal boiling point of 110.30°C and a boiling point elevation...
A certain liquid X has a normal boiling point of 110.30°C and a boiling point elevation constant =Kb0.71·°C·kgmol−1. A solution is prepared by dissolving some barium hydroxide (BaOH2) in 350.g of X. This solution boils at 110.7°C. Calculate the mass of BaOH2 that was dissolved.
Homework Answers
∆T = 110.7 - 110.30 = 0.4 °C
∆T = Kb * molality
= 0.4 = 0.71 * molality
Molality = 0.5634
Moles / mass of solvent ( in kg) = 0.5634
Moles / (350*10^-3) = 0.5634
Moles = 0.1972
Mass of Ba(OH)2 = mole * molar mass
= 0.1972* 171.34
= 33.79 grams
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