Question

100 mL of 0.1 M Al2(SO4)3 is mixed with 100 mL of 0.1 M AlCl3 solution. Calculate molarity of Al3+ ion if final solution has density 1.2 g/mL.

Answer 1

Answer:

100 mL of 0.1 M Al2(SO4)3 is mixed with 100 mL of 0.1 M AlCl3 solution. Calculate molarity of Al3+ ion if final solution has density 1.2 g/mL.100 mL of 0.1 M Al2(SO4)3 is mixed with 100 mL of 0.1 M AlCl3 solution. Calculate molarity of Al3+ ion if final solution has density 1.2 g/mL.

Explanation:

100 mL of 0.1 M Al2(SO4)3 is mixed with 100 mL of 0.1 M AlCl3 solution. Calculate molarity of Al3+ ion if final solution has density 1.2 g/mL .k

Answer 2

The number of moles of Al3+ ions in 100 mL of 0.1 M Al2(SO4)3 solution is:

moles of Al3+ ions in Al2(SO4)3 solution = 0.1 M × 0.1 L = 0.01 moles

Similarly, the number of moles of Al3+ ions in 100 mL of 0.1 M AlCl3 solution is:

moles of Al3+ ions in AlCl3 solution = 0.1 M × 0.1 L = 0.01 moles

When the two solutions are mixed, the total number of moles of Al3+ ions will be:

Total moles of Al3+ ions = moles of Al3+ ions in Al2(SO4)3 + moles of Al3+ ions in AlCl3

= 0.01 moles + 0.01 moles

= 0.02 moles

The total volume of the solution is 200 mL. The density of the final solution is 1.2 g/mL. Therefore, the mass of the final solution is:

Mass of solution = volume × density

= 200 mL × 1.2 g/mL

= 240 g

The molarity of Al3+ ions in the final solution can be calculated as follows:

Molarity of Al3+ ions = moles of Al3+ ions / volume of solution in liters

= 0.02 moles / (200 mL / 1000 mL/L)

= 0.1 M

Therefore, the molarity of Al3+ ions in the final solution is 0.1 M.