Calculate the molality of a solution containing 22.75 g of glycerol (C3H8O3) in 79.6 g of ethanol (C2H5OH).

3.11 mol/kg

Explanation:

Molality M = number of moles of solute, n/mass of solvent, m

To calculate the number of moles of glycerol (C₃H₈O₃) in 22.75 g of glycerol, we find its molar (molecular) mass, M',

So, M' = 3 × atomic mass of carbon + 8 × atomic mass hydrogen + 3 × atomic mass of oxygen

= 3 × 12 g/mol + 8 × 1 g/mol + 3 × 16 g/mol = 36 g/mol + 8 g/mol + 48 g/mol = 92 g/mol.

So, number of moles of glycerol, n = m'/M' where m' = mass of glycerol = 22.75 g and M' = molecular mass of glycerol = 92 g/mol

So, n = m'/M'

n = 22.75 g/92 g/mol

n = 0.247 mol

So, the molality of the solution M = n/m

Since m = mass of ethanol = 79.6 g = 0.0796 kg, substituting the value of n into the equation, we have

M = 0.247 mol/0.0796 kg

M = 3.11 mol/kg

So, the molality of the solution is 3.11 mol/kg.