Question

the molar concentration of chloride ion in the solution obtained by mixing 300ml of 3.0 M NaCl and 200 ml of 4.0 M Ba Cl₂is 5.0 M explain??

Answer 1

1 mole NaCl has 1 mole of chloride ions
1 mole of BaCl₂ has 2 moles of chloride ions
So, number of moles of chloride ions in 1 L of 3 M NaCl solution = 3 moles
Thus 300 mL of 3 M NaCl solution contain = $\frac{3}{1000}\times 300$ = 0.9 moles chloride ion
Number of chloride ions in 1 L of 4 M BaCl₂ = 8
Thus 200 mL of 4 M BaCl₂ solution contain = $\frac{8}{1000}\times 200$ = 1.6 moles chloride ion
Total moles of chloride ion in solution = 0.9 + 1.6 = 2.5 moles
Total volume of mixture = 300 + 200 = 500 ml = 0.5 L
Concentration of chloride ions in the mixture = number of moles/ volume in litres
= 2.5/0.5
=5 M