Question

No of ions present in 1ml of 0.1M calcium chloride sol is

Answer 1

0.1M solution means that you have 0.1 mol of CaCl2 in 1000mL solution 
But CaCl2 ↔ C+ 2+ + 2Cl- = 3 ions / molecule 
Therefore the 0.1M CaCl2 solution = 0.3M in respect of ions. 

You have 0.3 mol ions in 1000 mL solution 
You have 0.3/1000 = 0.0.0003 mol ions in 1 mL solution 

Number of ions: 
1 mol of ions = 6.022*10^23 ions 
0.0003 mol ions = (0.0003)*(6.022*10^23) = 1.81*10^20 ions.

Answer 2

Answer:

The required number of ions present in 1ml of 0.1M calcium chloride solution is 1.8 × 10²⁰ ions.

Explanation:

Molarity = $\frac{moles of solute}{volume of solution}$ × 1000

∴ moles of solute = $\frac{Molarity * Volume of solution(mL)}{1000}$

∴ Moles of ​Calcium Chloride(CaCl₂) = $\frac{0.1*1}{1000}$ = 10⁻⁴ moles

Number of molecules of CaCl₂ = Moles × Avogadro's no

= 10⁻⁴ × 6.022 × 10²³

= 6.022 × 10¹⁹ molecules

1 molecule of CaCl₂ gives 3 ions.

∴ No. of ions = 3 × number of molecules of CaCl₂​

= 3×6.022 × 10¹⁹

= 1.8 × 10²⁰ ions

Thus, there are two negative chloride ions for every positive calcium ion in this molecule. An ionic substance is neutral in charge since the positive and negative charges balance each other out.