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Chapter 12: Problem 161

Calculate the number of moles of each ion present in \(2.00 \times 10^{2} \mathrm{~cm}^{3}\) of (a) \(0.200 \mathrm{M} \mathrm{NaCl}\), (b) \(0.350 \mathrm{M} \mathrm{K}_{3} \mathrm{PO}_{4}\) (c) \(1.44 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

Short Answer

Expert verified
In summary, the number of moles of each ion present in the given solutions are: (a) Na+: 0.0400 moles, Cl-: 0.0400 moles (b) K+: 0.210 moles, PO43-: 0.0700 moles (c) Al3+: 0.288 moles, NO3-: 0.864 moles

Step by step solution

01

Find the moles for each solution

Firstly, we have to find the total moles of each compound using the formula: Moles = Volume × Molarity Note that the given volume is 2.00 × 10^2 cm³, which we need to convert to liters because Molarity is in moles per liter. \(1 L = 1000 cm^{3}\), so \((2.00 \times 10^{2} cm^{3}) = 0.200 L\). Now let's find the moles for each compound: (a) Moles of NaCl = Molarity × Volume = 0.200 M × 0.200 L = 0.0400 moles (b) Moles of K3PO4 = Molarity × Volume = 0.350 M × 0.200 L = 0.0700 moles (c) Moles of Al(NO3)3 = Molarity × Volume = 1.44 M × 0.200 L = 0.288 moles
02

Calculate moles for the ions of each compound

Now we will break each compound into its constituent ions to find the number of moles for each ion: (a) 1 mole of NaCl gives 1 mole of Na+ and 1 mole of Cl-. So, Moles of Na+: 0.0400 moles Moles of Cl-: 0.0400 moles (b) 1 mole of K3PO4 gives 3 moles of K+ and 1 mole of PO43-. So, Moles of K+: 3 × 0.0700 moles = 0.210 moles Moles of PO43-: 0.0700 moles (c) 1 mole of Al(NO3)3 gives 1 mole of Al3+ and 3 moles of NO3-. So, Moles of Al3+: 0.288 moles Moles of NO3-: 3 × 0.288 moles = 0.864 moles In summary, the number of moles of each ion present in the given solutions are: (a) Na+: 0.0400 moles, Cl-: 0.0400 moles (b) K+: 0.210 moles, PO43-: 0.0700 moles (c) Al3+: 0.288 moles, NO3-: 0.864 moles

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