Question

Calculate the number of Cl^-1 ion in 100ml of the solution which is formed by mixing of 200ml 2M NaCl ( aq) solution , 800ml 1 M MgCl2 (aq) solution and one litre 5.3 ×10^-4 M ccl4 solution

Answer 1

Answer:

Explanation:

Ccl4 does not releases Cl ions when in solution as it has non-planar structure.

While, MgCl2 , Nacl releases Cl ions in solution.

Thus, Nacl , Mgcl2 and Ccl4 will be in the ratio of 1 :4 :5

Since, in the question it is given for 100ml this will not affect the ratios because the mixture has uniform distribution in solution.

Nacl will have 10ml.

Mgcl2 will have 40ml.

Ccl4 will have 50ml.

Since, number of moles=molarity * volume

Now, number of moles of Mgcl2,

=1*40/100

=0.4 moles

Number of moles of Nacl,

=2*10/100

=0.2 moles

Since, the limiting factor is 0.2 moles hence, 1 of moles of Nacl releases 1 mole Cl ion hence, 0.2 mole Nacl releases 0.2 moles Cl ions.

1 of moles of Macl2 releases 2 mole Cl ion hence, 0.2 mole Nacl releases 0.4 moles Cl ions.

Thus, total Cl ions =0.2+0.4 = 0.6moles

Total Cl ions = 0.6* 6.022*10^22

= 3.6*10^22.