Match the following property.
A B
(i) Law of Multiple proportions. (p) Richter
(ii) Law of Combining volumes (q) Proust
(iii) Law of Reciprocal proportions. (r) GayLussac
(iv) Law of Constant composition. (s) Dalton
A. i - s, ii - p, iii - r, iv - q
B. i - s, ii - r, iii - p, iv - q
C. i - s, ii - r, iii - q, iv - p
D. i - q, ii - r, iii - p, iv - s
Answer : Option B
32. Two oxides of a metal ‘M’ contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is \(M_{3}O_{4}\) , find that of the second.
A. \(M_{2}O_{3}\)
B. \(M_{2}O\)
C. \(MO_{2}\)
D. \(M_{3}O_{2}\)
Answer : Option A
Explanation :
Let ‘x’ be the atomic mass of metal ‘M’
In the oixde \(M_{3}O_{4}\), the mass of ‘M’ = 72.4 and that of ‘O’ = 27.6
\(\therefore M_{\frac{72.4}{x}}O_{\frac{27.6}{16}}=M_{3}O_{4}\)
\(\therefore \frac{72.4}{x}:\frac{27.6}{16}=3:4\)
\(\therefore x=56\)
For second oxide, the mass of ‘M’ = 70 and that of ‘O’ =30
\(\therefore M_{\frac{70}{56}}O_{\frac{30}{16}}\)
\(=M_{1.25}O_{1.875}\)
\(=M_{1}O_{1.5}\ or\ M_{2}O_{3}\)
33. Naturally occuring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81)
A. 20% and 80%
B. 80% and 20%
C. 25% and 75%
D. 75% and 25%
Answer : Option A
Explanation :
Let the % of isotope with atomic mass 10.01 be ‘x’
\(\therefore \)% of isotope with atomic mass 11.01 = 100-x
⇒ Avg at mass = \(\frac{10.01x+(100-x)11.01}{100}=10.81(given)\)
34. Calculate the mass percent of Na and S in sodium sulphate.
A. Na = 16.2%, S = 22.54%
B. Na = 32.39%, S = 11.26%
C. Na = 22.54%, S = 32.39%
D. Na = 32.39%, S = 22.54%
Answer : Option D
35. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass.
36. Calculate the amount of carbon dioxide that can be produced when 1 mole of carbon is burnt in 16 g of dioxygen.
37. Calculate the concentration of nitric acid in moles per liter which has a density , 1.41 g/mL. %W/W of nitric acid is
A. 15.44M
B. 0.064M
C. 0.077M
D. 12.87M
Answer : Option A
Explanation :
→ 69% w/w means 100 g nitric acid soution contain 69 g of nitric acid by mass.
\(\therefore moles\ of\ HNO_{3}=\frac{69}{63}=1.095\)
→ Vol. of 100 g nitric acid solution = \(\frac{100}{1.41}=0.07092L\)
\(\therefore \) moles per litre = \(\frac{1.095}{0.07092}=15.44\)
38. In a reaction : \(N_{2(g)}+3H_{2(g)}\rightarrow 2NH_{3(g)}\), 2000g \(N_{2}\) reacts with 1000g \(H_{2}\) which reactant will left unreacted ? How much ?
A. \(N_{2},\ 2428g\)
B. \(H_{2},\ 428.6g\)
C. \(N_{2},\ 571.4g\)
D. \(H_{2},\ 571.4g\)
Answer : Option D
Explanation :
\(N_{2}+3H_{2}\rightarrow 2NH_{3}\)
28 g 6 g
2000 g (?)
\(=\frac{2000\times 6}{28}=428.6g\)
→ But we are given 1000 g \(H_{2}\) There fore 1000 - 4286 =571.4 g \(H_{2}\) will left
39. Calculate the number of sulphate ions in 100mL of 0.001M ammonium sulphate solution.
A. \(6.022\times 10^{-19}\)
B. \(6.022\times 10^{19}\)
C. \(6.022\times 10^{20}\)
D. \(6.022\times 10^{-20}\)
Answer : Option B
Explanation :
→ No of moles of \((NH_{4})_{2}SO_{4}\) = molarity x vol(L)
= 0.001 x 0.1 = 0.0001
\(\therefore \) No of \(SO_{4}^{2-}\) ions = \(0.0001\times 6.022\times 10^{23}\)
\(6.022\times 10^{19}\)
40. Calculate the molarity of a solution of ethanol in water in which mole fraction of ethanol is 0.040.
A. 2.31M
B. 0.213M
C. 0.0213M
D. 23.1M
Answer : Option A
Explanation :
\(\rightarrow X_{ETOH}=\frac{n_{(ETOH)}}{n_{(ETOH)}+n_{(H_{2}O)}}\)
\(\therefore 0.04=\frac{n_{(ETOH)}}{n_{(ETOH)}+55.55}\)
\(\therefore n_{(ETOH)}=2.31\)