Version A Consider the following for the next two (2) questions: A 1.0620g sampl
ID: 525122 • Letter: V
Question
Version A Consider the following for the next two (2) questions: A 1.0620g sample containing an unknown amount ofNa,c,o, was titrated against a 0.02007 M KMno4 solution. The titration required 36.65 mL ofthe KMno. solution to reach the endpoint. 12. How many moles of oxalate (czo3) are present in the sample? A. 2.188 x 10.3 C. 1.839 x 10.3 E. 5.227 x 104 D. 9.560 x 104 B. 1.030 x 103 13. What percentage by mass of the sample is oxalate (coi 88.0 gmol? E. 44.6% A. 15.2% B. 9.19% C, 5.05% D. 20.1% is and a Brensted-Lowry base is 14. By definition, a Bransted-Lowry acid A. an electron pair donor: an electron pair acceptor B. a solution with a pH 7; a solution with a pHExplanation / Answer
(12)
2 KMnO4 (aq.)+3 H2SO4(aq.)+5H2C2O4( aq.)--------> K2SO4(aq.)+2MnSO4(aq.)+10CO2(g)+8H2O(l)
Moles of KMnO4 = 0.02007 * 36.65 / 1000 = 0.0007356 mol
From the balanced equation,
moles of oxalate = 0.0007356 * 5 / 2 = 0.001839 mol
Therefore moles of oxalate = 1.839 * 10-3 mol
(C)
(13)
Mass of oxalate = 0.001839 * 88 = 0.1618 g.
% of oxalate = (0.1618/1.0620) * 100 = 15.23 %
(A)
(14) (E) Proton donor is an acid and proton acceptor is a base
(15)
(D) Amount of a substance deposited at an electrode is directly proportional to Quantity of electricity passed through it.
(16) (E) 6 electrons
3 Mg (s) + 2 Al3+ (aq.) ------------> 3 Mg2+ (aq.) + 2 Al (s)
(17) (A) 2 H+ (aq.) + 2 e ------------> H2 (g)
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