Answer:
Temperature of freezing for the solution is. 3.3°C
Explanation:
This is an excersise of coligative property (freezing low depression)
ΔT = Kf . m
Mass of solute (biphenyl) → 0.617 g
Mass of solvent (cycloexane) → 25 g
m = molality (mol of solute in 1kg of solvent)
Let's find out moles of solute:
Moles = Mass / Molar mass
Moles = 0.617 g / 154 g/m → 0.004 moles
To know molalilty, let's convert 25 g to kg
25 g = 0.025 kg
moles / kg = molality
0.004 m / 0.025 kg = 0.16m
ΔT = T°freezing pure solvent - T° freezing solution
6.5°C - T° freezing solution = 20°C/m . 0.16m
6.5°C - T° freezing solution = 3.20°C
6.5°C - 3.20°C = T° freezing solution → °3.3 °C
