71. Which one of the following statements is incorrect ?
A. All elements are homogeneous system
B. Compounds made up of a number of elements are heterogeneous.
C. A mixture is not always heterogeneous
D. Smoke is a heterogeneous mixture.
Answer : Option B
72. A balanced chemical equation is in accordance with
A. Avogadro’s law.
B. Law of constant proportions
C. Law of conservation of mass
D. Law of gaseous volumes.
Answer : Option C
73. The atomic weights of two elements X and Y are 20 and 40 respectively. If ‘a’ gm of X contains ‘b’ atoms, how many atoms are present in ‘2a’ gm of Y ?
A. b
B. a
C. 2b
D. a/2
Answer : Option A
Explanation :
No of moles of X = \(\frac{a}{20}\)
\(\therefore \) No of atoms of X = \(\frac{a}{20}\times N=b\)
\(\therefore a=\frac{20b}{N}\)
No of moles of Y = \(\frac{2a}{40}\)
\(\therefore \) No of atoms of Y = \(\frac{2a}{40}\times N\)
\(=\frac{2}{40}\times \frac{20b}{N}\times N=b\)
74. If the components of air are \(N_{2}\) ,78%, \(O_{2}\) , 21% ; Ar, 0.9% and \(CO_{2}\), 0.1% by volume, what will be the molecular weight of air ?
A. 28.9
B. 32.4
C. 16.4
D. 14.5
Answer : Option A
Explanation :
Mol wt of air = \(\frac{78\times 28+21\times 32+0.9\times 40+0.1\times 44}{78+21+0.9+0.1}\)
75. Calculate the molarity of a solution obtained by mixing 50mL of 0.5M \(H_{2}SO_{4}\) and 75 mL of 0.25M \(H_{2}SO_{4}\)
A. 0.375M
B. 0.35M
C. 0.045M
D. 0.45M
Answer : Option B
Explanation :
→ No.of moles of 0.05L \(H_{2}SO_{4}\) = 0.5 x 0.05 = 0.025
→ No.of moles of 0.075L \(H_{2}SO_{4}\) = 0.25 x 0.075 = 0.01875
\(\therefore \) Total no. of moles = 0.025 + 0.01875 = 0.04375
Total vol = 0.05L + 0.075L = 0.125L
\(\therefore \) Molarity = \(\frac{0.04375}{0.125}=0.35M\)
76. Which of the following has the highest normality ?
A. 1M \(H_{2}SO_{4}\)
B. 1M \(H_{3}PO_{3}\)
C. 1M \(H_{3}PO_{4}\)
D. 1M \(HNO_{3}\)
Answer : Option C
77. In an experiment, 4 gm of \(M_{2}O_{x}\) oxide was reduced to 2.8 gm of the metal. If the atomic mass of the metal is 56 gm/mol, the number of oxygen atoms in the oxide is (AFMC 2010)
A. 1
B. 2
C. 3
D. 4
Answer : Option C
Explanation :
→ 1 Mol \(M_{2}O_{x}\) = (2 x 56 + 16x) gm
→ Now, (2 x 56 +16x) gm of oxide = 112 gm metal
\(\therefore \) 4 gm of oxide = \(\frac{112\times 4}{112+16x}\) gm metal
→ But \(\frac{112\times 4}{112+16x}\) = 2.8 (given)
\(\therefore x=3\)
Match the following
Column - I Column - II
(i) femto (P) \(10^{9}\)
(ii) yotta (q) \(10^{-15}\)
(iii) giga (r) \(10^{-18}\)
(iv) atto (s) \(10^{24}\)
A. i - q, ii - p, iii - r, iv - s
B. i - s, ii - q, iii - p, iv - r
C. i - q, ii - s, iii - p, iv - r
D. i - r, ii - s, iii - p, iv - q
Answer : Option C
79. The total number of atoms of all elements present in mole of ammonium dichromate is
A. 19
B. \(6.023\times 10^{23}\)
C. \(114.47\times 10^{23}\)
D. \(84\times 10^{23}\)
Answer : Option C
Explanation :
→ Molecular formula of ammonium dichromate is \((NH_{4})_{2}Cr_{2}O_{7}\)
80. 0.32 gm of a metal on treatment with an acid gave 112 mL of hydrogen at STP. Calculate the equivalent weight of the metal
A. 58
B. 32
C. 11.2
D. 24
Answer : Option B
Explanation :
\(Eq\ wt\ of\ metal=\frac{wt\ of\ metal\times 11200}{vol\ of\ H_{2}\ in\ ml\ displaced\ at\ STP}\)