Modern family: Addition

State level Olympiad (6)

Routine was same for Chemistry exam at the weekend. Two differences were no. of questions and format of the questions.
Questions were objective in nature and so no is also increased from 5 to 50. Questions were as given below.

1. A student is asked to measure 30.0 g of methanol (d = 0.7914 g/mL at 25˚C) but has only a graduated cylinder with which to measure it. What volume of
methanol should the student use to obtain the required
30.0 g?

(A) 23.7 mL (B) 30.0 mL
(C) 32.4 mL (D) 37.9 mL

Sol. Let us use the equation for density to determine the volume.
d = m/V
V = m/d
Now, we can substitute m = 30.0 g and d = 0.7914 g/mL , we get
V = 30.0/0.7914 = 37.9 mL
Thus, the correct answer is D .

2. A flame test was performed to confirm the identity of a
metal ion in solution. The result was a green flame. Which of the following metal ions is indicated?

(A) copper (B) sodium
(C) strontium (D) zinc

Sol. This question relies on your knowledge of flame test colors. Copper is blue-green, sodium is yellow, stron-tium is red, and zinc is colorless. The answer is A .

3. When phenolphthalein is added to an aqueous solution
containing one of the following solutes the solution turns
pink. Which solute is present?

(A) NaCl (B) KC2H3O2
(C) LiBr (D) NH4NO3

Sol. The key fact to understand is that in basic solutions, phenolphtalein turns pink. Thus, we need to find a ba-
sic compound in order to create the pink solution. Both NaCl and LiBr are neutral salts, while NH4NO3 is weakly
acidic due to the presence of NH4+, conjugate acid of the weak base NH3. Thus, the correct answer is KC2H3O2 , or
B . This works due to the presence of a neutral ion (K+) and the weak base (C2H3O2−, conjugate base of the weak
acetic acid HC2H3O2).

4. Solid camphor is insoluble in water but is soluble in
vegetable oil. The best explanation for this behavior is
that camphor is a(n)

(A) ionic solid (B) metallic solid
(C) molecular solid (D) network solid

Sol. Ionic solids like NaCl are are often soluble in water, but not in nonpolar solvents such as oil. Network solids and metallic solids such as diamond and titanium are generally not soluble in either solvent. A molecular solid , such
as C , can be nonpolar (such as solid saturated fats) and therefore soluble in oil.

5. A student performed an experiment to determine the ratio of H2O to CuSO4 in a sample of hydrated copper(II)
sulfate by heating it to drive off the water and weighing
the solid before and after heating. The formula obtained
experimentally wasCuSO4 •5.5H2O but the accepted
formula is CuSO4 •5H2O. Which error best accounts for
the difference in results?

(A) During heating some of the hydrated copper(II)
sulfate was lost.
(B) The hydrated sample was not heated long enough to
drive off all the water.
(C) The student weighed out too much sample initially.
(D) The student used a balance that gave weights that
were consistently too high by 0.10 g.

Sol. Overheating the hydrate will cause the removal of some of the CuSO4, which will cause the mass percent of
H2O to appear bigger. The option that fits this analysis is A .

6. An aqueous solution is known to contain Ag+ , Mg2+, and
Sr2+ ions. Which reagent should be used to selectively
precipitate the Ag+ ?

(A) 0.20 M NaCl (B) 0.20 M NaOH
(C) 0.20 M Na2SO4 (D) 0.20 M Na3PO4

Sol. We need to find an anion that precipitates selectively with silver, but not magnesium or strontium.
• Na+ does not interact with any of the three cations.
• MgCl2 and SrCl2 are both soluble in water, but AgCl is not.
• OH− precipitates with both magnesium and silver.
• (SO4) 2− salts of all three metals are insoluble except for Mg2+
• (PO4) 3− salts of all three metals are insoluble.
Thus, the reagent that will most selectively precipitate the Ag+ is the 0.20 M NaCl , or A .

7. What is the coefficient for O2 when the following reaction
__As2S3 + __O2 → __ As2O3 + __SO2
is correctly balanced with the smallest integer coefficients?

(A) 5 (B) 6 (C) 8 (D) 9

Sol. We need to balance the S, so we can add a coefficient of 3 to the SO2.
As2S3 + O2 −−→ As2O3 + 3 SO2
Now to balance the oxygens, we have 9 O on the products side, and 2 O on the reactants side. Thus, the reaction is
As2S3 + 9/2 O2 −−→ As2O3 + 3 SO2

However, to eliminate the fractional coefficient, we multiply by 2 throughout the equation.
2 As2S3 + 9 O2 −−→ 2 As2O3 + 6 SO2
Thus, the coefficient on the O2 is 9 , or D .

8. Which compound contains the highest percentage of nitrogen
by mass?

(A) NH2OH (M = 33.0) (B) NH4NO2 (M = 64.1)
(C) N2O3 (M = 76.0) (D) NH4NH2CO2 (M = 78.1)

Sol. We are already given the molar masses, so we need to determine the number of N in each compound. For
Compound A, we have
(14.0/33.0) × 100% = 42.4%
(28.0/64.1) × 100% = 43.7%
(28.0/76.0) × 100% = 36.8%
(28.0/78.1) × 100% = 35.9%
Thus, the correct answer is NH4NO2 , or B .

9. How many neutrons are in 0.025 mol of the isotope (54, 24) Cr?

(A) 1.5 × 1022 (B) 3.6 × 1023
(C) 4.5 × 1023 (D) 8.1 × 1023

Sol. We can use dimensional analysis with Avogadro’s number.
(0.025 mol) × (6.02 × 10^23 atoms/mol) × (54 − 24) neutrons/atom
= 4.5 × 10^23 neutrons
Thus, the correct answer is C .

10. Magnesium chloride dissolves in water to form

(A) hydrated MgCl2 molecules
(B) hydrated Mg2+ ions and hydrated Cl–
ions
(C) hydrated Mg2+ ions and hydrated Cl2
2– ions
(D) hydrated Mg atoms and hydrated Cl2 molecules

Sol. MgCl2 is a typical ionic compound which dissolves in water. The dissociation of magnesium chloride with
water is as follows. The term ”hydrated” just means that each ion is surrounded by water molecules, represented by
”aq” in the equation above)
MgCl2(s) −−→ Mg2+(aq) + 2 Cl−(aq)
This reaction corresponds to answer B .

11. Enzymes convert glucose (M= 180.2) to ethanol (M= 46.1)
according to the equation
C6H12O6 → 2C2H5OH + 2CO2
What is the maximum mass of ethanol that can be made
from 15.5 kg of glucose?

(A) 0.256 kg (B) 0.512 kg
(C) 3.96 kg (D) 7.93 kg

Sol. This is a stoichiometry problem, in which we want the mass of the ethanol formed from a given amount of glucose.
(15.5 kg C6H12O6) ×
(1 mol C6H12O6 /180 g C6H12O6)
× (2 mol C2H5OH /1 mol C6H12O6)
× (46.1 g C2H5OH /1 mol C2H5OH) = 7.93 kg

Thus, the correct answer is D .

12. Commercial vinegar is a 5.00% by mass aqueous solution
of acetic acid, CH3CO2H (M = 60.0). What is the molarity
of acetic acid in vinegar? [density of vinegar = 1.00 g/mL]

(A) 0.833 M (B) 1.00 M
(C) 1.20 M (D) 3.00 M

Sol. We can solve this with dimensional analysis.
(5.00/100) × (1 mol/60.0 g) ×
(1.00 g/mL) × (1000 mL/L)
= 0.833 M
Thus, the answer is A .