Q13.90P - Chemistry: Molecular Nature Of Matter And Change

Question

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

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{MKBr}\mathbf{ } \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{065}{\mathbf{MHNO}}_{\mathbf{3}}\mathbf{ } \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{MKHSO}}_{\mathbf{4}}\mathbf{ }\mathbf{ }\mathbf{ } \\ \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{Methanol}\left(C_2{H}_5\mathrm{OH}\right) \end{gathered}$$

Step-by-Step Solution

Answer

Answer 

 

1. 0.6 mol of solute particles are present in $0.3M KBr$.
2. 0.13 mol of solute particles are present in$0.065M HN{O}_3$
3. $2\times {10}^{-4} mol$ of solute particles are present in  ${10}^{-4}M KHS{O}_4$
4. 0.06 mol of solute particles are present in  $0.06M C_2{H}_{5}OH$

 

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

$\mathbf{molarity}\mathbf{=}\frac{\mathbf{molesofsolute}}{\mathbf{L} \mathbf{of} \mathbf{solution}}$,  

Step 2: moles of solute in 0.3 M KBr

 

 consists of 2 particles for each $KBr\left(K^{+}andB{r}^{-}ions\right)$.

 $\left[0.3molKBr/L\right]\left[2molparticle/1molKBr\right]\left(1L\right)=0.6molpfparticles$

Step 3: moles of solutes in 0.065 M HNO 3

 

 consists of 2 particles for each mole of molecule (hydronium and nitrate ions 

 $\left[0.065mol/L\right]\left[2molparticle/1molHN{O}_3\right]\left(1L\right)=13.0molofparticles$

Step 4: moles of solutes in 10 - 4 MKHSO 4

 

$KHS{O}_4$ consists of 2 particles for each mole of molecule (sulphate and potassium ions)

 _{$\left[{10}^{-4}mol KHS{O}_4/L\right]\left[2molparticles/1molKHS{O}_4\right]\left(1L\right)=2_{\times }{10}^{-4}molofsoluteparticles$}

 

Step 5: moles of solutes in 0 . 06 Methanol C 2 H 5 OH

 

Ethanol is not an ionic compound .so, it does not dissociate into ions but remain as single particle. So, no. of moles of particles is same as the number of moles of molecules 

$\left[0.06mol{C}_2{H}_{5}OH/L\right]\left[1molparticles/1mol{C}_2{H}_{5}OH\right]\left(1L\right)=0.06molofsoluteparticles$