Question

Question 25 of 35 > What is molarity of A1+ ions in a solution prepared by dissolving 1.06 g of Al2(SO4)3 in 250,0 mL of water. Molar mass of Al2(SO4)3 is 342.15 g/mol. Enter your answer in the provided box. For this question, show your complete work in Part 2

Answer 1

The molecular weight of the Al2(SO4)2 =342.15 gm/mol

Molarity of Al2(SO4)2 =

M = (1.06 * 1000) / (342.15 * 250) = 0.01239 moles

A solution of Al2(SO4)2 generate a 2 moles Al3+ of and 3 moles of SO4--

therefor a molarity of the Al3+ ion =

M (Al3+) = 0.01239 * 2 = 0.02478 moles

Molarity of stock solution of Al3+ = 0.02478 moles

The molecular weight of the KIO3 =214.001 gm/mol

Molarity of KIO3 =

a quantity of solute KIO3 is 5.86 mole

therefor total quantity is = 5.86 * 214.001 = 1254.04 gm

and volume of solution 2.80 L = 2800 ml

then calculate the

M = (1254.04 * 1000) / (214.01 * 2800) = 2.0927 moles

Molarity of stock solution of KIO3 = 2.0927 moles

Molarity of dilute solution of KIO3 =

calcualting using below formula

M1 * V1 = M2 * V2

M1= Initial molarity = 2.0927

V1 = Initial Volume = 10 ml

M2 = Final Molarity = ?

V2 = Final volume = 100 ml

therefor M2 = (M1 * V1)/ V2

M2 = ( 2.0927 * 10 ) / 100

M2 = 0.20927 mole