The molarity of 5.80 g of LiNO₃ in 505 mL of solution is 0.167 mol/L
Given mass of solute LiNO₃ = 5.80 g
Volume of solution = 505 mL = 0.505 L
Molar mass of LiNO₃ = 68.946 g/mol
Therefore, Number of moles in 5.80 g LiNO₃ = 5.80 / 68.946 = 0.0841 mol
Hence, Molarity of the solution = Number of moles of solute / Volume of the solution
= 0.0841 / 0.505 = 0.167 mol/L
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