Question

calculate the molality and molarity of a solution formed by dissolving 27.8 g of LiI in 500 mL of water

Answer 1

Molar mass of LiI,
MM = 1*MM(Li) + 1*MM(I)
= 1*6.968 + 1*126.9
= 133.868 g/mol


mass(LiI)= 27.8 g

use:
number of mol of LiI,
n = mass of LiI/molar mass of LiI
=(27.8 g)/(133.868 g/mol)
= 0.2077 mol

1)
Assuming density of water = 1 g/mL
m(solvent)= 500 g
= 0.500 kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(0.2077 mol)/(0.500 Kg)
= 0.415 molal
Answer: 0.415 molal

2)
volume = 500 mL
= 0.500 L

use:
Molality,
m = number of mol / volume of solution in L
=(0.2077 mol)/(0.500 L)
= 0.415 M
Answer: 0.415 M