Chemistry - Mole Concept Question with Solution | TestHub

Molar mass of BeCl₂ = $9+71=80g/mol$

Moles of BeCl₂ = $\frac{0.8}{80}=0.01$

Therefore, Molarity = $\frac{MolesofBeC{l}_2}{Volumeofsolution(mL)}\times 1000=\frac{0.01}{500}\times 1000=0.02M$

Now, take 20 mL of 0.02 M solution and dilute it 10 times. So, the final volume = $20\times 10=200mL$.

Hence, the final molarity of this solution can be calculated as:

$$M_1{V}_1=M_2{V}_2$$

$$0.02M\times 20mL=M_2\times 200mL$$

$M_2=\frac{0.02\times 20}{200}=\frac{0.4}{200}=0.002M=2\times 10^{-3}M$

BeCl₂ → Be²⁺ (aq) + 2Cl⁻ (aq)

2×10⁻³ M – –

- 2×10⁻³ M (2×2×10⁻³) M

[Cl⁻] = 4 × 10⁻³ M

In 2 mL solution (2/1000 L solution)

∴ Moles of Cl⁻ = Molarity × volume in litere

= 4 × 10⁻³ × 2/1000

= 8 × 10⁻⁶ moles