Molar mass of LiBr,
MM = 1*MM(Li) + 1*MM(Br)
= 1*6.968 + 1*79.9
= 86.868 g/mol
mass(LiBr)= 97.7 g
use:
number of mol of LiBr,
n = mass of LiBr/molar mass of LiBr
=(97.7 g)/(86.87 g/mol)
= 1.125 mol
volume , V = 7.5*10^2 mL
= 0.75 L
use:
Molarity,
M = number of mol / volume in L
= 1.125/0.75
= 1.50 M
Answer: 1.50 M
Question 8 (2 points) Determine the molarity of a solution formed by dissolving 97.7 g LiBr...
ASAP
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