Answer:
0.271 M NO₃⁻
Explanation:
To find the molarity of the nitrate ion (NO₃⁻), you need to (1) convert grams to moles (via molar mass), then (2) convert moles Al(NO₃)₃ to moles NO₃⁻, then (3) convert mL to L, and then (4) calculate the molarity. When (Al(NO₃)₃) dissolves in water, it dissociates into 3 nitrate ions. The final answer should have 3 sig figs.
(Steps 1 + 2)
Molar Mass (Al(NO₃)₃): 26.982 g/mol + 3(14.007 g/mol) + 9(15.998 g/mol)
Molar Mass (Al(NO₃)₃): 212.985 g/mol
1 Al(NO₃)₃ = 1 Al³⁺ and 3 NO₃⁻
6.25 g Al(NO₃)₃ 1 mole 3 moles NO₃⁻
------------------------- x ----------------- x ----------------------- = 0.0880 moles NO₃⁻
212.985 g 1 mole Al(NO₃)₃
(Steps 3 + 4)
325.0 mL / 1,000 = 0.3250 L
Molarity = moles / volume
Molarity = 0.0880 moles / 0.3250 L
Molarity = 0.271 M
