Question

calculate the molality of 1.22m glucose. density 1.22g/ml

Answer 1

consider 1 L of solution
volume = 1 L = 1000 mL

mass fo solution = density * volume
= 1.22 g/mL*1000 mL
= 1220 g

Now use :
Molarity = number of mol of glucose / volume of solution in L
1.22 = number of mol of glucose / 1 L
number of mol of glucose = 1.22 mol

Molar mass of C6H12O6,
MM = 6*MM(C) + 12*MM(H) + 6*MM(O)
= 6*12.01 + 12*1.008 + 6*16.0
= 180.156 g/mol

mass of C6H12O6,
m = number of mol * molar mass
= 1.22 mol * 180.156 g/mol
= 219.8 g

so,
mass of solvent = mass of solution - mass of glucose
= 1220 g - 219.8 g
= 1000.2 g
= 1.0002 Kg

molality = number of mol of glucose / mass of solvent in Kg
= 1.22 mol / 1.0002
= 1.22 molal

Answer: 1.22 molal