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Question
Calculate the molarity of the following solution:
30 g of \[\ce{Co(NO3)2 * 6H2O}\] in 4.3 L of solution.
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Solution
Molar mass of \[\ce{Co(NO3)2 * 6H2O}\]
= 58.7 + 2(14 + 48) + 6 × 18 g mol−1
= 58.7 + 2(62) + 108 g mol−1
= 58.7 + 124 + 108 g mol−1
= 290.7 g mol−1
∴ Number of moles of \[\ce{Co(NO3)2 * 6H2O}\] = `"Mass"/"Molar mass"`
= `(30 "g")/(290.7 "g mol"^(-1))`
= 0.103 mol
Volume of solution = 4.3 L
Molarity of solution = `"Number of moles of solute"/"Volume of solution in L"`
= `(0.103 "mol")/(4.3 "L")`
= 0.024 M
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