Question

Sucrose (C12H22O11), a nonionic solute, dissolves in water (normal freezing/melting point 0.0°C) to form a solution. If some unknown mass of sucrose is dissolved in 150g of water and this solution has a freezing/melting point of -0.56°C, calculate the mass of sucrose dissolved. Kfp for water is 1.86°C/m. [must show work including units to receive credit].

Answer 1

Given that Kfp for water is 1.86°C/m

depression in freezing point = normal freezing/melting point 0.0°C - freezing/melting point of solution

= 0.0 °C - (- 0.56°C )

= 0.56 °C

depression in freezing point = i*m*Kf

for nonionic solute ; i = 1

First calculate the molality of sucrose as follows:

0.56 °C= 1 * m * 1.86°C/m

m = 0.30 mole / kg

mass of water = 150 g or 0.150 kg

now calculate the number of moles of sucrose as follows:

molality = number of moles / solvent inKG

number of moles = 0.30 mole / kg *0.150 kg

= 0.045 mole sucrose

Molar mass of Sucrose is 342.2965 g/mol.

amount of sucrose in g = number of mole * molar mass

= 0.045 mole sucrose *342.2965 g/mol

= 15.40 g sucrose