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TS EAMCET Chemistry - Matter and Laws of Chemical Combination Study Materials with Answer

Matter and Laws of Chemical Combination

Exam Duration: 45 Mins Total Questions : 30

1 )

The solid like conducting state of gases with free electrons is called

(a)
sol
(b)
gel
(c)
plasma
(d)
All of these

Solution :

Solid like conducting state of gases is called plasma.

2 )

which of the following statements about a compound is incorrect?

(a)
A molecule of a compound has atoms of different elements
(b)
A molecule of a compound has atoms of different elements
(c)
A molecule of a compound has atoms of different elements
(d)
A molecule of a compound has atoms of different elements

3 )

In which of the following numbers all zeroes are significant?

(a)
30.000
(b)
0.700
(c)
0.00050
(d)
0.0030

Solution :

30.000 has all significant zeroes.

4 )

The answer of the calculation \(\frac{2.568\times5.8}{4.168}\) in significant figures will be

(a)
3.570
(b)
3.6
(c)
3.57
(d)
3.579

Solution :

\(\frac{2.568\times5.8}{4.168}=3.5735\). Thus, in significant figures=3.6

5 )

Number of significant figures in 78.000 g, 0.0206 g and 3.002 g are

(a)
3, 4, 5
(b)
2, 5, 4
(c)
3, 3, 4
(d)
5, 3, 4

Solution :

Number	Significant figures
78.000	5
0.0206	3
3.002	4

6 )

Law of constant composition does not hold true for

(a)
endothermic compounds
(b)
exothermic compounds
(c)
stoichiometric compounds
(d)
non-stoichiometric compounds

Solution :

Law of constant composition does not hold true for non-stoichiometric compounds.

7 )

Formation of CO and CO₂ illustrates the law of

(a)
conservation of mass
(b)
multiple proportion
(c)
reciprocal proportion
(d)
constant proportion

Solution :

Formation of CO and CO₂ illustrates the law of multiple proportion.

8 )

Which of the following reactions is not correct according to the law of conservation of mass?

(a)
2Mg(s) + O₂(s) \(\rightarrow\) 2MgO(s)
(b)
C₃H₈(g) + O₂ (g) \(\rightarrow\) CO₂(g) + H₂O(l)
(c)
P₄(s) + 5O₂(g) \(\rightarrow\) P₄O₁₀(s)
(d)
CH₄(g) + 2O₂(g) \(\rightarrow\) CO₂(g) + 2H₂O(g)

Solution :

C₃H₈(g) + O₂ (g) \(\rightarrow\) CO₂(g) + H₂O(l) is not correct.
Correct equation is
C₃H₈(g) + 5O₂ (g) \(\rightarrow\) 3CO₂(g) + 4H₂O(l)

9 )

One part of element A reacts with two parts of another elements B. 6 parts of element C reacts with 4 parts of element B. If A and C combine together, the ratio of their weights will be governed by

(a)
law of conservation of mass
(b)
law of reciprocal proportions
(c)
law of definite proportions
(d)
law of multiple proportions

Solution :

Ratio will be governed by law of reciprocal proportions.

10 )

Insulin contains 3.4% sulphur. What will be the minimum molecular weight of insulin?

(a)
94.117
(b)
1884
(c)
941.176
(d)
- 976

Solution :

For minimum molecular mass, there must be one S-atom per insulin molecule.
If 3.4 g of S is present, the molecular mass = 100
If 32 g of S is present, the molecular mass will be
\(=\frac{100\times32}{3.4}=941.176\)

11 )

Molecular weight of a tribasic acid is W. Its equivalent weight will be

(a)
W/2
(b)
W
(c)
W/3
(d)
3W

Solution :

Equivalent weight = \(\frac{Molecular \ \ weight}{Basicity}=\frac{W}{3}\)

12 )

Number of molecules in 4.25 g of NH₃ is

(a)
1.505 x 10²³
(b)
6.02 x 10²³
(c)
3.01 x 10²³
(d)
None of these

Solution :

17 g of NH₃ contains = 6.023 x 10²³ molecules of NH₃
4.25 g NH₃ will contain = \(\frac{6.023\times10^{23}}{17}\times4.25\)
= 1.505 x 10²³ molecules of NH₃

13 )

Which of the following weight the most?

(a)
One mole of water
(b)
One g-atom of nitrogen
(c)
One mole of sodium
(d)
One molecule of H₂SO₄

Solution :

One mole of water=18 g, 1 g atom of nitrogen = 14 g
1 mole of sodium = 23 g
1 molecule of H₂SO₄ = \(\frac{98}{6.023\times10^{23}}g=1.627\times10^{-22}g\)
Hence, one mole of sodium weight the most.

14 )

Compounds having same empirical formula always have same

(a)
law of gaseous volumes
(b)
Avogadro's hypothesis
(c)
Dalton's atomic theory
(d)
Berzelius hypothesis

Solution :

Compounds having same empirical formula always have same percentage composition by mass.

15 )

A sample of AlF₃ contains 3.0 x 10²⁴ F^- ions. The number of formula units of this sample are

(a)
9.0 x 10²⁴
(b)
3.0 x 10²⁴
(c)
0.75 x 10²⁴
(d)
1.0 x 10²⁴

Solution :

3F^- \(\equiv\) 1 Formula unit(AlF₃)
3.0 x 10²⁴ F^- = 1 x 10²⁴ Formula units (AlF₃)

16 )

The equivalent mass of Cl is 35.5 and the atomic mass of Cu is 63.5. The equivalent mass of copper chloride is 99. Hence, formula of copper chloride is

(a)
CuCl
(b)
Cu₂Cl
(c)
CuCl₂
(d)
None of these

Solution :

Equivalent mass of copper chloride = 99
Equivalent mass of chlorine = 35.5
Equivalent mass of copper = 99 - 35.5 = 63.5
Valency of copper = \(\frac{Atomic \ \ mass \ \ of \ \ copper}{Equivalent \ \ mass \ \ of \ \ copper}=1\)
Formula of copper chloride is CuCl.

17 )

Which of the following pairs contain equal number of atoms?

(a)
22.4 L (STP) of nitrous oxide and 22.4 L of nitric oxide
(b)
1 millimole of HCl and 0.5 millimole of H₂ S
(c)
1 mole of H₂O₂ and 1 mole of N₂O₄
(d)
11.2 cc (STP) of nitrogen and 0.015g of nitric oxide

Solution :

Number of atoms in N₂ = \(\frac{11.2\times10^{-3}\times6.023\times10^{23}\times2}{22.4}=6.023\times10^{20}\)
Number of atoms in NO = \(\frac{0.015\times2\times6.023\times10^{23}}{30}=6.023\times10^{20}\)

18 )

Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g

Student

Readings

(i)

3.01

3.05

(ii)

2.99

2.95

On the basis of given data, mark the correct option out of the following statement.

(a)
Results of both the students are neither accurate nor precise
(b)
Results of student A are both precise and accurate
(c)
Result of student B are neither precise nor accurate
(d)
Both are inaccurate scientifically

Solution :

Results of student B are neither precise nor accurate

19 )

Which of the following statements indicates that law of multiple proportions is being followed?

(a)
Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1 : 2
(b)
Carbon forms two oxides namely CO₂ and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2 : 1
(c)
When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed
(d)
At constant temperature and pressure, 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour

Solution :

Law of multiple proportions is followed in the formation of two oxides of carbon CO₂ and CO, where mass of oxygen which combine with fixed mass of carbon are in the simple ratio 2 : 1

20 )

Calculate the number of moles left after removing 10²¹ molecules from 200 mg of CO₂

(a)
0.00576
(b)
0.0034
(c)
0.00288
(d)
None of the above

Solution :

Molecular mass of CO₂=12+2x16=44
44g of CO₂ contains=6.023x10²³ molecules
ஃ 0.2g of CO₂ contains=\(\frac{6.023\times10^{23}\times0.2}{44}\)
=0.02738x10²³ molecules
∵ Molecules removed=1x10²¹
ஃ Number of molecules remained=0.02738x10²³-1x10²¹
=2.738x10²¹-1x10²¹=1.738x10²¹
=\(\frac{1.738\times10^{21}}{6.023\times10^{23}}\)mol=0.00288 mol

21 )

Vapour density of a gas is 11.2. Volume occupied by 2.4 g of this at STP will be

(a)
2.4 L
(b)
2.24 L
(c)
22.4 L
(d)
11.2 L

Solution :

Molecular mass = 2 x vapour density = 2 x 11.2 = 22.4 g
Number of moles of gas = \(\frac{2.4}{22.4}\)
1 mole occupies = 22.4 L volume
\(\frac{2.4}{22.4}\) mole will occupy = 22.4 x \(\frac{2.4}{22.4}\) = 2.4 L

22 )

The relative abundance of two isotopes of atomic weight85 and 87 is 75% and 25% respectively. The average atomic weight of element is

(a)
40.0
(b)
75.5
(c)
85.5
(d)
86.0

Solution :

Average atomic mass = \(\frac{RA(1) \times at.mass(1)+RA(2)\times at.mass(2)}{RA(1)+RA(2)}\)
Average atomic weight = \(\frac{75\times85+25\times87}{75+25}=\frac{6375+2175}{100}=\frac{8550}{100}=85.50\)

23 )

0.75 moles of a solid A₄ are heated with 2 moles of O₂ (g) in a sealed vessel.Completely using up the reactants, the reaction produces only one compound. It is observed that when the temperature is used to that of initial temperature, the contents of the vessel exhibit a pressure equal to 1/2 of the original pressure. The formula for the product will be

(a)
A₂O₃
(b)
A₃O₈
(c)
A₃O₄
(d)
AO₂

Solution :

Since, both the reactants are completely consumed, therefore, the ratio of stoichiometric coefficients would be 0.75 : 2 or 3:8.
Thus, 3A₄ + 8 O₂ \(\rightarrow\) Product
If final pressure is that of half of oxygen initially, the molecular formula will be A₃O₄ , therefore,
3A₄ + 8 O₂ \(\rightarrow\) 4A₃O₄

24 )

400 mg of iron capsule contains 100 mg of ferrous fumarate (CHCOO)₂ Fe. The percentage of iron present in it is approximately

(a)
8.2 %
(b)
14 %
(c)
25 %
(d)
33 %

Solution :

Molecular mass of (CHCOO)₂Fe = 170 g mol^-1
100 mg of (CHCOO)₂ Fe contains = \(\frac{56}{170}\times100 \ \ mg\ \ Fe=32.9 \ \ mg \ \ of \ \ Fe\)
32.9 mg of Fe is present in 400 mg of Fe capsule
% of Fe in capsule = \(\frac{32.9}{400}\times100=8.2\) %

25 )

A certain amount of a metal whose equivalent mass is 28, displaces 0.7 L of H₂ at STP from an acid. The mass of an element would be

(a)
1.75 g
(b)
6.875 g
(c)
3.50 g
(d)
7.00 g

Solution :

Mass of hydrogen = \(\frac{0.7}{22.4}\times2= 0.0625 \ \ g \)
0.0625 g of H is displaced by = x g of metal
1 g of metal is displaced by \(\frac{x}{0.0625} \ \ g \ \ of \ \ metal\)
\(\frac{x}{0.0625} =28\)
Equivalent weight of metal, x = 28 x 0.0625 = 1.75 g

26 )

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecule is

(a)
1 : 4
(b)
7 : 32
(c)
1 : 8
(d)
3 : 16

Solution :

\(\frac { { n }_{ { O }_{ 2 } } }{ { n }_{ { N }_{ 2 } } } =\frac { { m }_{ { O }_{ 2 } } /{ M }_{ { O }_{ 2 } } }{ { m}_{ { N }_{ 2 } } /{ M}_{ { N }_{ 2 } }} =\frac{{ m }_{ { O }_{ 2 } }}{{ m }_{ { N}_{ 2 } }}=\frac{1}{4}\times\frac{28}{32}=\frac{7}{32}\)

27 )

How many moles of magnesium phosphate Mg₃(PO₄)₂ will contain 0.25 mole of oxygen atoms?

(a)
0.02
(b)
3.125 x 10^-2
(c)
1.25 x 10^-2
(d)
2.5 x 10^-2

Solution :

8 moles of O-atoms are contained in 1 mole of Mg₃(PO₄)₂ Hence, 0.25 mole of O-atoms are contained in moles of Mg₃(PO₄)₂=1/8 x 0.25 = 3.125 x 10^-2

28 )

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit. The mass of one mole substance will

(a)
be a function of molecular mass of the substance
(b)
remains unchanged
(c)
increase two fold
(d)
decreases twice

29 )

Number of atoms in 560 g of Fe (atomic mass 56 mol^-1) is

(a)
twice that of 70 g N
(b)
half that of 20 g H
(c)
Both (a) and (b)
(d)
None of the above

Solution :

In 560 g of Fe, number of moles = \(\frac{560 \ \ g}{56 \ \ g \ \ mol^{-1}}=10 \ \ mol\)
In 70 g of N, number of moles = \(\frac{70 \ \ g}{14 \ \ g \ \ mol^{-1}}=5 \ \ mol\)
20 g of H = \(\frac{20 \ \ g}{1 \ \ g \ \ mol^{-1}}=20 \ \ mol\)

30 )

In an organic compound of molar mass 108 g mol^-1, C, H and N present in 9 : 1 : 3.5 by weight. Molecular formula can be

(a)
C₆H₈N₂
(b)
C₇H₁₀N
(c)
C₅H₆N₃
(d)
C₄H₁₈N₃

Solution :

Element	%	Atomic mass	Relative number of moles	Simple ratio
C	9	12	0.75	3
H	1	1	1	4
N	3.5	14	0.25	1

ஃ Empirical formula=C₃H₄N and empirical formula mass=36+4+14=54
n=\(\frac{108}{54}\)=2
ஃ Molecular formula=(C₃H₄N)₂=C₆H₈N₂

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