JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 23

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Multiple Choice Questions

Question 1: One fermi is equal to
(a) 10−13 cm
(b) 10−15 cm
(c) 10−10 cm
(d) 10−12 cm
Show Answer & Explanation

Answer: (a) 10−13 cm

Explanation:
1. 1 fermi (fm) = 10−15 m.
2. 1 m = 100 cm, so 10−15 m = 10−13 cm.

Teacher's Note:
1. The fermi is the unit used for nuclear sizes. Convert metres to centimetres by multiplying by 100.

Question 2: The prefix 1018 is called
(a) Giga
(b) Nano
(c) Mega
(d) Exa
Show Answer & Explanation

Answer: (d) Exa

Explanation:
1. Giga = 109, Mega = 106, Nano = 10−9.
2. 1018 is Exa.

Teacher's Note:
1. Learn the SI prefixes in steps of 103. Exa follows Peta (1015).

Question 3: The numbers 161 cm, 0.161 cm and 0.0161 cm contain how many significant figures respectively?
(a) 3, 4 and 5
(b) 3, 3 and 3
(c) 3, 3 and 4
(d) 3, 4 and 4
Show Answer & Explanation

Answer: (b) 3, 3 and 3

Explanation:
1. Non-zero digits are always significant.
2. Zeros before the first non-zero digit only fix the decimal point and are not significant.
3. All three numbers contain the digits 1, 6, 1, so each has 3 significant figures.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 4: The number of significant figures in 0.00051 is
(a) 5
(b) 3
(c) 2
(d) 4
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Leading zeros are not significant.
2. Only 5 and 1 count, so there are 2 significant figures.

Teacher's Note:
1. Moving the decimal point never changes the count of significant figures. Write the number as 5.1 × 10−4 to see this clearly.

Question 5: Which of the following halogens can be purified by sublimation?
(a) F2
(b) Cl2
(c) Br2
(d) I2
Show Answer & Explanation

Answer: (d) I2

Explanation:
1. Iodine passes directly from solid to vapour on gentle heating.
2. The vapour condenses back to pure solid on a cold surface, leaving non-volatile impurities behind.
3. F2 and Cl2 are gases and Br2 is a liquid at room temperature.

Teacher's Note:
1. Sublimation works for solids such as iodine, camphor, naphthalene and ammonium chloride. The impurity must not sublime.

Question 6: The number of significant figures in the final answer of \( \frac{(29.2-20.2)(1.79\times10^{5})}{1.37} \) is
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. 29.2 − 20.2 = 9.0. In subtraction the answer keeps the least number of decimal places, so it is written 9.0 (two significant figures).
2. In multiplication and division the result carries as many significant figures as the least precise factor.
3. The least precise term is 9.0 with 2 significant figures, so the answer has 2.

Teacher's Note:
1. Apply the addition/subtraction rule first, then the multiplication/division rule.

Question 7: A 81.4 g sample of ethyl alcohol contains 0.002 g of water. The mass of pure ethyl alcohol, expressed to the proper number of significant figures, is
(a) 81.398 g
(b) 71.40 g
(c) 81.4 g
(d) 81 g
Show Answer & Explanation

Answer: (c) 81.4 g

Explanation:
1. Mass of pure alcohol = 81.4 − 0.002 = 81.398 g.
2. In subtraction the result is limited by the least number of decimal places, which is one (81.4).
3. So the answer is 81.4 g.

Teacher's Note:
1. Never report more decimal places than the least precise measurement allows.

Question 8: The unit J Pa−1 is equivalent to
(a) m3
(b) cm3
(c) dm3
(d) None of these
Show Answer & Explanation

Answer: (a) m3

Explanation:
1. Joule = N m and Pascal = N m−2.
2. J Pa−1 = (N m)/(N m−2) = m3.

Teacher's Note:
1. Reducing a unit to base units is the quickest way to identify it.

Question 9: From the following masses, the one which is expressed nearest to the milligram is
(a) 16 g
(b) 16.4 g
(c) 16.428 g
(d) 16.4284 g
Show Answer & Explanation

Answer: (c) 16.428 g

Explanation:
1. One milligram is 0.001 g, the third decimal place.
2. 16.428 g is given to exactly the third decimal place, while 16.4284 g goes beyond a milligram.

Teacher's Note:
1. Match the last digit of a reading with the smallest unit asked for.

Question 10: The number of significant figures in \( 6.02\times10^{23} \) is
(a) 23
(b) 3
(c) 4
(d) 26
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. In scientific notation only the digit term 6.02 is counted.
2. It has 3 significant figures.

Teacher's Note:
1. The power of ten never contributes to the count of significant figures.

Question 11: The prefix zepto stands for
(a) 109
(b) 10−12
(c) 10−15
(d) 10−21
Show Answer & Explanation

Answer: (d) 10−21

Explanation:
1. 109 is giga, 10−12 is pico, 10−15 is femto.
2. Zepto is 10−21.

Teacher's Note:
1. Femto, atto, zepto, yocto are 10−15, 10−18, 10−21, 10−24.

Question 12: The number of significant figures in 6.0023 is
(a) 5
(b) 4
(c) 3
(d) 1
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. Zeros lying between two non-zero digits are significant.
2. All five digits 6, 0, 0, 2, 3 are significant.

Teacher's Note:
1. Captive (sandwiched) zeros always count.

Question 13: The number of significant figures in 60.0001 is
(a) 5
(b) 6
(c) 3
(d) 2
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. The zeros are between non-zero digits 6 and 1, so they are significant.
2. All six digits count.

Teacher's Note:
1. Compare with 0.0001, where the zeros are leading and not significant.

Question 14: Which pair of substances illustrates the law of multiple proportions?
(a) CO and CO2
(b) H2O and D2O
(c) NaCl and NaBr
(d) MgO and Mg(OH)2
Show Answer & Explanation

Answer: (a) CO and CO2

Explanation:
1. The law of multiple proportions needs two elements forming more than one compound.
2. CO and CO2 contain C and O: for a fixed 12 g of carbon the masses of oxygen are 16 g and 32 g, in the ratio 1 : 2.

Teacher's Note:
1. Look for the same two elements in two different compounds.

Question 15: 1.0 g of an oxide of A contains 0.5 g of A. 4.0 g of another oxide of A contains 1.6 g of A. The data illustrate the law of
(a) Reciprocal proportions
(b) Constant proportions
(c) Conservation of energy
(d) Multiple proportions
Show Answer & Explanation

Answer: (d) Multiple proportions

Explanation:
1. First oxide: 0.5 g A combines with 0.5 g O, so 1 g A combines with 1 g O.
2. Second oxide: 1.6 g A combines with 2.4 g O, so 1 g A combines with 1.5 g O.
3. Ratio of oxygen masses = 1 : 1.5 = 2 : 3, a simple whole-number ratio.

Teacher's Note:
1. Always compare the masses of one element that combine with a fixed mass of the other.

Question 16: A chemical equation is balanced according to the law of
(a) Multiple proportions
(b) Reciprocal proportions
(c) Conservation of mass
(d) Definite proportions
Show Answer & Explanation

Answer: (c) Conservation of mass

Explanation:
1. Atoms are neither created nor destroyed in a reaction.
2. So the number of atoms of each element must be equal on both sides of the equation.

Teacher's Note:
1. Balancing equations is a direct application of Lavoisier's law.

Question 17: Carbon and oxygen form two oxides in which carbon : oxygen by mass is 12 : 16 and 12 : 32. These figures illustrate the
(a) Law of multiple proportions
(b) Law of reciprocal proportions
(c) Law of conservation of mass
(d) Law of constant proportions
Show Answer & Explanation

Answer: (a) Law of multiple proportions

Explanation:
1. For the same 12 g of carbon, the masses of oxygen are 16 g and 32 g.
2. 16 : 32 = 1 : 2, a simple whole-number ratio, which is the law of multiple proportions.

Teacher's Note:
1. Fix the mass of one element, then compare the other.

Question 18: A sample of calcium carbonate has the composition Ca = 40%, C = 12%, O = 48%. If the law of constant proportions holds, the mass of calcium in 4 g of CaCO3 obtained from another source will be
(a) 0.016 g
(b) 0.16 g
(c) 1.6 g
(d) 16 g
Show Answer & Explanation

Answer: (c) 1.6 g

Explanation:
1. By the law of constant proportions, every pure sample of CaCO3 has 40% calcium.
2. Mass of Ca = 40/100 × 4 g = 1.6 g.

Teacher's Note:
1. Pure compounds have the same percentage composition no matter how they are prepared.

Question 19: n g of substance X reacts with m g of substance Y to form p g of R and q g of S. The reaction X + Y → R + S shows that
(a) n − m = p − q
(b) n + m = p + q
(c) n = m
(d) p = q
Show Answer & Explanation

Answer: (b) n + m = p + q

Explanation:
1. Mass of reactants = n + m and mass of products = p + q.
2. By the law of conservation of mass, n + m = p + q.

Teacher's Note:
1. Mass is conserved overall, not individually for each substance.

Question 20: In compound A, 1.00 g of nitrogen unites with 0.57 g of oxygen. In compound B, 2.00 g of nitrogen combines with 2.24 g of oxygen. In compound C, 3.00 g of nitrogen combines with 5.11 g of oxygen. These results obey the law of
(a) Constant proportions
(b) Multiple proportions
(c) Reciprocal proportions
(d) Dalton's law of partial pressures
Show Answer & Explanation

Answer: (b) Multiple proportions

Explanation:
1. Oxygen per 1.00 g of nitrogen: A = 0.57 g, B = 1.12 g, C = 1.703 g.
2. Ratio 0.57 : 1.12 : 1.703 ≈ 1 : 2 : 3, a simple whole-number ratio.
3. So the law of multiple proportions is followed.

Teacher's Note:
1. Reduce every compound to the same mass of one element before comparing.

Question 21: Hydrogen combines with oxygen to form H2O in which 16 g of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form CH4 in which 2 g of hydrogen combine with 6 g of carbon. If carbon and oxygen combine together, they will show the ratio of
(a) 6 : 16 or 12 : 32
(b) 6 : 18
(c) 1 : 2
(d) 12 : 24
Show Answer & Explanation

Answer: (a) 6 : 16 or 12 : 32

Explanation:
1. Fixed 2 g of hydrogen combines with 16 g of oxygen and with 6 g of carbon.
2. So C : O = 6 : 16 (or a simple multiple, e.g. 12 : 32), by the law of reciprocal proportions.

Teacher's Note:
1. In the law of reciprocal proportions the third ratio is the same as, or a simple multiple of, the ratio of the two elements that combined with a common element.

Question 22: 2 g of hydrogen combine with 16 g of oxygen to form water, and with 6 g of carbon to form methane. In carbon dioxide, 12 g of carbon are combined with 32 g of oxygen. These figures illustrate the law of
(a) Multiple proportions
(b) Constant proportions
(c) Reciprocal proportions
(d) Conservation of mass
Show Answer & Explanation

Answer: (c) Reciprocal proportions

Explanation:
1. The ratio C : O from the two hydrogen compounds is 6 : 16 = 3 : 8.
2. In CO2 the ratio is 12 : 32 = 3 : 8, the same value.
3. This agreement is the law of reciprocal proportions.

Teacher's Note:
1. Two elements that each combine with a third show the same ratio when they combine with each other.

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