Question

The freezing point of water H₂O is 0.00°C at 1 atmosphere.

How many grams of barium sulfide (169.4 g/mol), must be dissolved in 297.0 grams of water to reduce the freezing point by 0.400°C ?

g barium sulfide.

Answer 1

delta T = Kf*molality

where, delta T = depression in freezing point

0.400 = 1.84 * molality

molality = 0.217 m

molality = number of moles of solute / mass of solution in kg

0.217 m = number of moles of solute / 0.297 kg

number of moles of solute = 0.217*0.297 = 0.0644 mole

number of moles of solute = mass of solute / molar mass of solute

0.0644 mole = mass of solute / 169.4 g.mol^-1

mass of solute = 10.9 g

Therefore, the mass of barium sulfide dissolved must be 10.9 g