Question

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is The mole fraction of Cl- in this solution is __________ M.

Answer 1

Strategy :

Here we are given mass of CaCl2, mass of water, and density of water. In order to calculate mol fraction of Cl- in the solution we must know total and moles of Cl-. So first step is to calculated moles of CaCl2 and by using it moles of Cl- are calculated by using mole ratio of CaCl2 as CaCl2 : Cl- ( 1 : 2 )

Once we get moles of Cl- then we can calculated moles of Ca²⁺ and water as well by using their molar masses.

Solution:

Step 1^st :

Calculation of moles of CaCl2

n CaCl2 = 23.7 g CaCl2 x ( 1 mol CaCl2/ 110.984 g CaCl2)

= 0.2135 mol CaCl2

Step 2:

Calculation of moles of Cl-

n Cl- = n CaCl2 x ( 2 mol Cl-/1 mol CaCl2)

=0.2135 mol CaCl2 x ( 2 mol Cl-/1 mol CaCl2)

= 0.4271 mol Cl-

Step 3 :

Calculation of moles of water

n H₂O = 375 g H2O x 1 mol H2O / 18.0148 g H2O

= 20.82 mol H2O

Step 4 :

Calculation of total moles

n Total = n Ca²⁺ + n H2O + n Cl-   ……….( since n CaCl2 = n Ca²⁺ ; mol ratio is 1 : 1 )

                =0.2135+20.816+0.4271

                =21.457 mol

Step 5:

Calculation of mol fraction of Cl

Mol fraction of Cl- = n Cl- / Total moles

                = 0.4271 / 21.457

                =0.02

Answer: Mole fraction of Cl- = 0.02