How many moles of LiOH are required to prepare 1.50L of 5.4 M LiOH?
1 answer:
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The molality of the solution = 17.93 m
<h3>Further explanation</h3>Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
- mass of water
6 L water = 6 kg water(ρ= 1 kg/L)
- molality
