Chemistry - Concentration Terms Question with Solution | TestHub

Molar mass of BeCl₂ = $9+71=80g/mol$

$$\begin{array}{rl}& \Rightarrow \text{ Moles of }{\mathrm{BeCl}}_2=\frac{0.8}{80}=0.01\\ & \therefore \text{ Molarity }=\frac{\text{ Moles of }{\mathrm{BeCl}}_2}{\text{ Volume of solution }(\mathrm{mL})}\times 1000\\ & =\frac{0.01}{500}\times 1000=0.02\mathrm{M}\end{array}$$

Now, take 20 mL of 0.02 M and dilute it 10 times. So, the final volume $=20\times 10=200mL$. Hence, the final molarity of this solution can be calculated as:

$$\begin{array}{rl}& \therefore {\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2\\ & \Rightarrow 0.02\times 20=200\times {\mathrm{M}}_2\text{ (final molarity) }\\ & \Rightarrow {\mathrm{M}}_2=2\times 10^{-3}\mathrm{M}\\ & \Rightarrow {\mathrm{BeCl}}_2\to {\mathrm{Be}}^{+2}(\mathrm{aq})+2{\mathrm{Cl}}^{-}(\mathrm{aq})\\ & 2\times 10^{-3}\mathrm{M}-\\ & -2\times 10^{-3}\mathrm{M}\left(2\times 2\times 10^{-3}\right)\mathrm{M}\end{array}$$

$[C{l}^{-}]=4\times 10^{-3}M$

in 2 mL solution $(\frac{2}{1000}\mathrm{L}$ solution).

∴    Moles of Cl^- = Molarity × volume in litre

$$=4\times 10^{-3}\times \frac{2}{1000}$$

$=8\times 10^{-6}moles$