Question

What is the osmolarity of 0.250g of CaCO3 that has been dissolved in 100.0 mL of H2O? (Please list steps for practice)

Answer 1

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 0.250 g

we have below equation to be used:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(0.25 g)/(100.09 g/mol)

= 2.498*10^-3 mol

volume , V = 100.0 mL

= 0.1 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.498*10^-3/0.1

= 2.498*10^-2 M

1 CaCO3 dissociates into 2 particle (1 Ca2+ and 1 CO32-)

So,

Osm = 2*M

= 2*2.498*10^-2 M

= 5.00*10^-2 osM

Answer: 5.00*10^-2 osM