Q13-91P - Chemistry: Molecular Nature Of Matter And Change

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 

Step-by-Step Solution

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. $8.0\times {10}^{-5}$ moles of solute particles are present in data-custom-editor="chemistry" $0.08M\mathit{ }pyridine\left(C_5{H}_{5}N\right)$ 

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

Step 1: 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,  $Molarity = \frac{(moles of solute)}{(L of solution) }$

Step 2: moles of solute in 0 . 02 M   C u S O 4

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.

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

  $2.0\times {10}^{-5} moles 2=4.0\times {10}^{-5}moles of solute particles.$

Step 3: moles of solute in 0 . 004 M   B a ( O H ) 2

 Ba(OH)₂ consists 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.

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

$4.0\times {10}^{-6} moles 3=1.2\times {10}^{-5 }moles of solute particles.$

Step 4: moles of solute in 0 . 08 M   P y r i d i n e   ( C 5 H 5 N )

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

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

$8.0\times {10}^{-5} moles1= 8.0\times {10}^{-5 }moles of solute particles.$

Step 5: moles of solute in 0 . 05 M   ( N H 4 ) 2 C O 3

$(N{H}_4{)}_{2}C{O}_3$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.

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

$5.0\times {10}^{-5} moles 3= 1.5\times {10}^{-4}moles of solute particles.$