Question

Number of chloride ions present in 100 ml of 0.1 M CaCl₂ solution is​

Answer 1

Given: Volume of CaCl₂ solution, V = 100ml

            Molarity of CaCl₂ solution, M = 0.1M

To Find: Number of chloride ions, N.

Solution:

To calculate N, the formula used:

• Molarity = number of moles(n)/ volume of solution ( in L)
• M = n / V
• In mL, M = (n /V) x 1000

Applying the formula :

0.1 = ( n/ 100) x 1000

0.1 = n x 10

n = 0.1/10

n = 1/100

n = 0.01 mol

As in CaCl₂, there are two chloride ions

So, the number of chloride ions = 2 x n x Avagardo's number (Nₐ)

here, Nₐ = 6.02 x 10²³ ions

N = 2 x 0.01 x 6.02 x 10²³

   = 0.02 x 6.02 x 10²³

   = 0.1204 x 10²³

   = 12.04 x 10²⁵ ions

N = 12.04 x 10²⁵ ions

Hence, the Number of chloride ions present in 100 ml of 0.1 M CaCl₂ solution is​ 12.04 x 10²⁵ ions.

Answer 2

Answer:

The number of chloride ions present in $CaCl_{2}$ solution measured is $1.2\times 10^{22}$.

Explanation:

Given,

The molarity of $CaCl_{2}$ solution, M = $0.1M$

The volume of the solution, V = $100 ml$ = $0.1 L$

The number of chloride ions present in $CaCl_{2}$ solution =?

As we know,

• Molarity = $\frac{Number of moles}{Volume}$
• Number of moles = $Molarity \times volume$
• Number of moles = $0.1\times 0.1$ = $0.01 mol$

Now, the dissociation of $CaCl_{2}$ occurs in the way given below:

• $CaCl_{2} \rightarrow Ca^{2+} + 2Cl^{-}$

From this reaction, we can say that one mole of $CaCl_{2}$ yields $2 moles$ of chloride ions.

As we know,

• 1 mole of $CaCl_{2}$ = $6.022\times 10^{23}$ molecules of $CaCl_{2}$
• $0.01$ moles of $CaCl_{2}$ = $0.01 \times 6.022\times 10^{23}$ = $0.6022\times 10^{22}$ molecules of $CaCl_{2}$.

Now, as we know, one mole of $CaCl_{2}$ yields $2 moles$ of chloride ions.

Therefore,

• The number of chloride ions present in $CaCl_{2}$ solution = $2\times 0.6022\times 10^{22}$ = $1.2\times 10^{22}$ chloride ions.