Question

How many moles of solute particles are present in 1 mL of each of the following aqueous solutions?

(a)0.02M CuSO₄

(b)0.004 MBa(OH)₂

(c)0.08 M pyridine (C₅H₅N)

(d) 0.05 M (NH₄)₂CO₃

Short answer

1. $4.0\times {10}^{-5}$moles of solute particles are present in data-custom-editor="chemistry" $0.02MCuS{O}_4$.
2. $1.2\times {10}^{-5}$moles of solute particles are present in data-custom-editor="chemistry" $0.004MBa(OH{)}_2$.
3. role="math" localid="1663315547524" $8.0\times {10}^{-5}$moles of solute particles are present in data-custom-editor="chemistry" $0.08M\mathit{}pyridine\left(C_5{H}_{5}N\right)$

4. role="math" localid="1663315581073" $1.5\times {10}^{-4}$moles of solute particles are present in $0.05M(N{H}_4{)}_{2}C{O}_3$

Step-by-step solution

Formula

To count the no. of moles of solute particles in a solution of an ionic compound,

First count the ions per mole and multiply by the no. of moles in solution.

For a covalent compound, the number of particles is equal to the number of molecules.

Also, $Molarity=\frac{(molesofsolute)}{(Lofsolution)}$

moles of solute in0.02M CuSO4

CuSO₄ consists of 2 particles for each mole of molecule because when you dissolve 1 mole CuSO₄ in solvent it dissolves into 1 mol copper ions and 1 mole sulphate ions, which gives you twice as many moles of solute particles.

$$\begin{gathered} Molarity=0.02M\times 0.001L \\ =2.0\times {10}^{-5}moles \end{gathered}$$

$2.0\times {10}^{-5}moles2=4.0\times {10}^{-5}molesofsoluteparticles.$

moles of solute in 0.004M Ba(OH)2

Ba(OH)₂consists of 3 particles for each mole of molecule because when you dissolve 1 mole Ba(OH)₂ in solvent it dissolves into 2 mol hydroxyl ions and 1 mole barium ions, which gives you thrice as many moles of solute particles.

$$\begin{gathered} Molarity=0.004M\times 0.001L \\ =4.0\times {10}^{-6}moles \end{gathered}$$

$4.0\times {10}^{-6}moles3=1.2\times {10}^{-5}molesofsoluteparticles.$

moles of solute in 0.08M Pyridine (C5H5N)

Pyridine consists of 1 particle for each mole of molecule because it does not dissociate into ions when dissolves in solvent.

$$\begin{gathered} Molarity=0.08M\times 0.001L \\ =8.0\times {10}^{-5}moles \end{gathered}$$

$8.0\times {10}^{-5}moles1=8.0\times {10}^{-5}molesofsoluteparticles.$

moles of solute in 0.05M (NH4)2CO3

$(N{H}_4{)}_{2}C{O}_3$consists of 3 particles for each mole of molecule because when you dissolve 1 mole$(N{H}_4{)}_{2}C{O}_3$ in solvent it dissolves into 1 mol carbonate ions and 2 mole ammonium ions, which gives you thrice as many moles of solute particles.

^{$$\begin{gathered} Molarity=0.05M\times 0.001L \\ =5.0\times {10}^{-5}moles \end{gathered}$$}

$5.0\times {10}^{-5}moles3=1.5\times {10}^{-4}molesofsoluteparticles.$