Question

Calculate the number of cl-and ca2+ions in 222g anhydrous CaCl2?


Take avagadro no be N
options :-
a) 4N, 2N
b) 2N, 4N
c) 1N, 2N
d) 2N, 1N

Answer 1

111g of cacl2 contain 35.5g of cl and 40g of Ca.then 222g of CaCl2 will contain 80g of Ca and 71g of . to calculate no. of ions we simply find out no. of moles and multiply it by avogadro number.

80/40*6.022*10power23 = 12.044*10power23 Ca2+ions

71/35.5*6.022*10power23 = 12.044*10power23 Cl-ions

Answer 2

Answer: The correct answer is Option a.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Given mass of calcium chloride = 222 g

Molar mass of calcium chloride = 110.98 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium chloride}=\frac{222g}{110.98g/mol}=2mol$

The chemical formula for calcium chloride is $CaCl_2$

In 1 mole of calcium chloride, 2 moles of $Cl^-$ ions are present and 1 mole of $Ca^{2+}$ ions are present.

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ (or N) number of atoms.

So, in 2 moles of calcium chloride, (2 × 2 × N) = 4N number of $Cl^-$ ions are present and (2 × 1 × N) = 2N number of $Ca^{2+}$ ions are present.

Hence, the correct answer is Option a.