Question number 1. You dissolve 329. mg of a compound (A w B x C y D z ) in wate

Question

Question number 1.

You dissolve 329. mg of a compound (AwBxCyDz) in water producing 19.79 mL of solution and use the photometer and standard solutions to find that the concentration of the A is 1.75 x 10-2 M. The molar mass of A is 233.65 g/mol.

Moles of A in the solution sample:________________________________

Mass of A in the solution sample: _________________________________

Mass % of A in the compound: _______________________________

Moles of A in 100 grams of the compound: ________________________________

Question number 2.

The compound, AwBxCy, like all compounds has an oxidation number of 0 (zero). The oxidation number of A is +1, the oxidation number of B is +2, and the oxidation number of C is

Explanation / Answer

1) moles of A = Molarity x vol (in L) = 1.75 x 10^-2 x 19.79/1000 =0.0003463 = 3.463 x 10^-4

2) mass of A = moles x mol wt = 0.0003463 x 233.65 = 0.081 gm

3) % of A = ( 100 x 0.081/0.329) = 24.6 %

4) for 100 gm compound moles of A = ( 100 x 0.0003463 / 0.329) = 0.105

second question

A;C = 2:3 then charge ratio is +2 : -3 , hence +4 : -6 now we need +2 to balance charge to zero i.e one B ( since charge of B is +2)

hence compound is A4BC6

3) moles of K = 8.24/39.098 = 0.211

moles of AL = 5.69/26.98 = 0.211

Al :K = 1:1

now for 100 gm compound

8.24+5.69 + 9y) + 18(z) = 100 ( 96 is mol wt of SO4 , 18 is mol wt of H2O)

96y + 18z = 86.07

as per charge Alis +3 while K is +1 so when K;Al = 1:1 SO4 moles = 4 times K

hence y = 0.211 x 4 = 0.822

z = 0.398

now K:Al: SO4 : H2O = 0.211:0.211:0.822:0.398

ratios is K:Al:SO4:H2O = 1:1:4:0.5

compound formula is K2Al2(SO4)8(H2O)

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