Question

1) An analysis indicated that the concentration of mercury in a sample of drinking water was 0.680 parts per billion. What quantity of mercury (g) was present in 15.0 L of the water, the density of which is 0.998 g/ml? A) 6.82 B) 1.72 x 10-8 C) 6.07 x 10-3 D) 1.06 x 10-6 E) 3.45 x 10-4 F) 4.54 x 10-5 G) 1.02 x 10-5 2) Boron has two naturally occurring isotopes, Boron-10 (10.01293 amu), and Boron-11 (11.0093 amu), with an average atomic mass of 10.811 amu. What is the % composition of the lighter isotope? a) 10.4 b) 19.9 c) 80.1 d) 22.6 e) 7.49 f) 6.22 g) 72.3 h) 26.7 3) Give the formulas for the following: a) gold(III) sulfate b) potassium phosphate c) propane d) ammonium hypochlorite e) diphosphorous pentoxide 1 12

Answer 1

1. We know, 1000 ppb = 1 mg/L ppb = parts per billion

Or, 1 ppb = 1/1000 mg/L

Or, 0.680 ppb = 1*0.680/1000 mg/L = 6.8 * 10^-4 mg/L = 6.8 * 10^-4 * 10^-3 g/L [1mg = 10^-3 g]

= 6.8 * 10^-7g/L

So, 1 L solution contains 6.8 * 10^-7 g mercury

Or, 15 L solution contains 6.8 * 10^-7 * 15 g mercury

= 1.02 * 10^-5 g

So answer is option G

2. Let mass of B¹⁰  = M₁ = 10.01293 amu

mass of B¹¹ = M₂ = 11.0093 amu

Mass of isotope = M = 10.811

Let % of B¹⁰  = x, % of B¹¹ = 100-x as total abundance = 100%

So total mass = [mass of isotope * their %] / total %

So, M = [M₁*x + M₂*(100-x)] / 100

Or, 10.811 = [10.01293*x + 11.0093(100-x)] / 100

Or, 10.811*100 = 10.01293*x + 1100.93 - 11.0093*x

Or, 1081.1 - 1100.93 = 10.01293x-11.0093x

Or, - 19.83 = - 0.99637x

Or, x = 19.83 / 0.99637 = 19.90

So % of lightest isotope = option b) 19.9

3.a. gold(III)sulphate = Au₂(SO₄)₃  [here valency of gold(Au) is 3 and sulphate is 2, which are exchanged to maintain charge neutrality]

b. potassium phosphate = K₃PO₄ [valency of K=1, phosphate = 3]

c. propane = CH₃CH₂CH₃

d. ammonium hypoclorite = NH₄OCl [both NH₄ and OCl has valecy 1]

e. diphosphorous pentoxide = P₂O₅ [2 P and 5O as named]