What is the freezing point of a solution that contains 36.0 g of glucose (c6h12o6) in 500.0 g of water? (kf for water is 1.86 c/m. the molar mass of glucose if 180.0 g per mole.)

Freezing point depression is a colligative property, which means that in only depends on the solute concentration.

The equation that relates freezing point depression with the solute concentration is:

ΔT = Kf * m

Where, ΔT is the freezing point depression, Kf is the cryoscopic constant and m is the molality or molal concentration.

Kf = 1.86 °C/m

m  = moles of solute / Kg of solvent

moles of solute = mass / [molar mass] = 36.0 g / 180.0 g/mol = 0.200 mol

Kg of solvent = 500.0 g / 1000 g/kg = 0.5000 Kg of water  

=> m = 0.200 / 0.5000 = 0.400 m

=> ΔT = 1.86 °C/m * 0.400m = 0.744 °C

Givent that the freezing point of pure water is 100°C, the new frezzing point is 100°C - 0.744 °C = 99.256°C.

99.256 °C   

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.