Problem 71-74. A 2.000 g sample of a Ni-Tl-Zn alloy is dissolved in nitric acid
ID: 1022631 • Letter: P
Question
Problem 71-74.
A 2.000 g sample of a Ni-Tl-Zn alloy is dissolved in nitric acid and 1.750 g of thallium(I) iodide, TIL is precipitated by the addition of hydriodic acid. Calculate the percentage of thallium in the alloy. (A) 39.03% (B) 53.98% (C) 62.22% (D) 87.50% A piece of silver foil having a mass of 0.5840 g had become tarnished because of formation of Ag_2 S, with a mass increase of 0.0010 g. What percentage of the original silver has been converted to Ag_2 S? (A) 0.6% (B) 1.2% (C) 1.8% (D) 2.4% What is the percentage of nitrogen by mass in (NH_4)_3 PO_4? (A) 14/62 times 100% (B) 21/80 times 100% (C) 14/113 times 100% (D) 42/149 times 100% A 2.00 g mixture of NaCl and NaNO_3 dissolved in water required 90 mL of 0.10 M AgNO_3 to precipitate all the chloride. What is the mass percentage of NaCl in the original sample? (A) 10 (B) 26 (C) 58 (D) 90Explanation / Answer
71.B
moles TlI = grams/molar mass. =1.750/331.3 = 0.00528 moles
moles Tl = same since there is 1 Tl in 1 TlI.
grams Tl = molesTl x molar mass Tl = 0.00528*204.4 = 1.079 g
%Tl= (mass Tl/mass sample)*100 =1.079/2 * 100 = 53.98%
72.C
The mass increase of 0.001 g can only be due to S.
0.001 g of S = 0.00003125 moles of S.
The S is present as Ag2S in which there are two moles of Ag per mole of S.
Therefore for every 0.00003125 moles of S, 0.0000625 moles of Ag have reacted.
So 0.00675 g silver have reacted from the original 0584 g.
The percentage of silver that has reacted is therefore 0.00675/0.584x100 = 1.16 or 1.2, answer c.
73.D
mass of the compound =149.086741 g/mol
mass of nitrogen = 42/149 * 100
74.B
millimoles of NaCl required = 0.10*90/1000 =0.009 M
so weight of NaCl required = 0.009*58.5 = 0.5265g
mass percentage of NaCl in the original sample = 0.5265/2 *100 = 26.30%