# Ion ,mole concentration precipitate silver ions questions ?

Aqueous XO4^-3 ion forms a precipitate with aqueous Ag ion.When 41 cm^3 of solution of silver aqueous ions of concentration of 0.2 mol/dm^3

were added to the solution then 1.2 gram of precipitate is formed

a) Calculate the mole of silver ion used and

b) calculate the amount of mole of Precipitate formed

c)Find Molar mass of Precipitate d)Determine the RAM of X and identify X .

Relevance

a)

XO4{3-} + 3 Ag{+} → Ag3XO4(s)

(0.041 dm^3) x (0.2 mol Ag{+}/dm^3) = 0.0082 mol Ag{+}

b)

(0.0082 mol Ag{+}) x (1 mol Ag3XO4(s) / 3 mol Ag{+}) = 0.0027 mol Ag3XO4(s)

c)

(1.2 g Ag3XO4(s)) / ((0.0082 mol Ag{+}) x (1 mol Ag3XO4(s) / 3 mol Ag{+})) =

439 g/mol

d)

(439 g Ag3XO4/mol) - ((107.8682 g Ag/mol) x 3) - ((15.99943 g O/mol) x 4) =

51.40 g X / mol

That atomic mass is half-way between V and Cr.

The CrO4 anion has a charge of 2-, so it does not fit this question.

The VO4 anion has a charge of 3-, so V is the answer to this question.

• 3Ag+ + XO4^-3 = Ag3(XO4)

moles of Ag+ ni the reaction =   0.2 mol/dm^3 * 0.041 dm^3 = 0.0082 moles

from the equation we see

moles of ppte = 0.0082 moles / 3 = 0.002733

mass of precipitate = 1.2g so molar mass of ppte= 1.2 g / .002733 moles

molar mass = 439.0 g / mol of precipitate

looking at 1 mole of Ag3(XO4)

atomic mass of Ag = 107.86 g/ mol

so 3 * 107.86 + X + 4*16 =  439.0 g / mol

323.58 +X+ 64 = 439.0

X = 439.0 - 64- 323.58 =  51.42 ATM

Chromium ( 52.00)