Question

50ml of 0.1M Hcl is mixed with 50ml of 0.05M h2so4 then the resultant [cl-] in the solution

Answer 1

Hey Dear,

◆ Answer -

[Cl-] = 0.05 M

◆ Explanation -

Volume of both HCl & H2SO4 is 50 ml = 0.05 L.

No of moles of Cl- ions in HCl solution -

Moles of Cl- = 0.05 × 0.1

Moles of Cl- = 0.005 mol

Resultant concentration of [Cl-] in the mixture -

Concentration = Moles of Cl- / Total volume

[Cl-] = 0.005 / (0.05+0.05)

[Cl-] = 0.05 M

Therefore, resultant [Cl-] in the solution is 0.05 M.

Hope you understand..

Answer 2

Answer:

Hey your answer is 0.05 M

Explanation:

Cl is present only in HCl

Now we know that number of moles of solute = Molarity × volume

Now nummber of moles of cl = 0.05L × 0.1M = 0.005 moles

Now when it is mixed with $H_{2} SO_{4\\}$ only volume increases . therefore Molarity becomes = $\frac{Moles of cl}{New volume}$ = $\frac{0.005}{0.1}$ = 0.05m

Hence the molarity comes out to be 0.05M