Question

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

Answer 1

The formula mass of Ca2+ is 40g while that of Cl- is 35.5g hence the formula mass of CaCl2 is 111g( 40+(35.5*2)) since there are two ions of cl- in CaCl2.
If 111g =6.022*10^23ions
Therefore 222g of CaCl2=(222*6.022*10^23)/111
Ans=1.205*10^24
There are 1Ca2+ in one unit of CaCl2 hence there are 1.205*10^24ions.
Similarly there are 2Cl- in one unit of CaCl2 hence there are( 1.205*10^24/2)=6.03*10^23 ions.
The number of Ca2+ is 1.205*10^24ions while that of Cl- is 6.03*10^23ions present in 222g of CaCl2.

Answer 2

The chemical formula and molar mass of calcium chloride are respectively CaCl2 and 111 g respectively. Thus, there are 6.022 X 1023 formula units in 111g of CaCl2. So,

number of formula units of CaCl2 in 111 g of CaCl2 = 6.022 X 1023

number of formula units of CaCl2 in 1g of CaCl2 = (6.022 X 1023) / 111

number of formula units in 222g of CaCl2 = (6.022 X 1023 X 222) / 111

= 12.044 X 1023

Now, each formula unit of CaCl2 contains one Ca2+ ion and two Cl- ions. Therefore

number of Ca2+ ions in one unit of CaCl2 = 1

number of Ca2+ ions in 12.044 X 1023 units of CaCl2 = 1 X 12.044 X 1022 = 1.2044 X 1024

Similarly

number of Cl- ions in one unit of CaCl2 = 2

number of Cl- ions in 12.044 X 1023 units of CaCl2 = 2 X 12.044 X 1023

= 2.4088 X 1024

Hence, there are 1.2044 X 1024 Ca2+ ions and 2.4088 X 1024 Cl- ions in 222 g of CaCl2.

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