Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Dimensions of pressure are same as that of

A
Energy

B
Force

C
Energy per unit volume

D
Force per unit volume

Solution

Pressure =\(\frac{Force}{Area}\)

\([MLT^{-2}]\over{[L^2]}\) = \([ML^{-1}T^{-2}]\)

\([ML^2T^{-2}]\over{[L^3]}\) = \([ML^{-1}T^{-2}]\)

Energy per unit volume

\([ML^2T^{-2}]\over{[L^3]}\) = \([ML^{-1}T^{-2}]\)
Question 2
1 / -0

Significant figures in 0.00051 are

A
5

B
3

C
2

D
4

Solution

Significant figures in 0.00051 are 2
Question 3
1 / -0

Which of the following elements of matter would best convey that there is life on earth?

A
Oxygen

B
Hydrogen

C
Carbon

D
Iron

Solution

Carbon is a chemical element with the symbol C and atomic number 6. It is non-metallic and tetravalent—making four electrons available to form covalent chemical bonds. It belongs to group 14 of the periodic table
Question 4
1 / -0

The number of significant figures in 60.0001 is

A
5

B
6

C
3

D
2

Solution

All the zeroes between two non-zero digit are significant. Hence in 60.0001 significant figures is 6.
Question 5
1 / -0

In known elements, the maximum number is of

A
Metals

B
Non-metals

C
Metalloids

D
None of these

Solution

Metals is the maximum number of known elements.
Question 6
1 / -0

ng of substance X reacts with mg of substance Y to form pg of substance R and qg of substance S. This reaction can be represented as, X + Y = R + S. The relation which can be established in the amounts of the reactants and the products will be.

A
n - m = p − q

B
n + m =p + q

C
n = m

D
p = q

Solution

law of conservation of mass.

\(\underset{ng}{X}+\underset{mg}{Y}\rightleftharpoons\,\underset{pg}{R}+ \underset{qg}{S}\,n\,+\,m\) = \(p\,+\,q\)
Question 7
1 / -0

Equivalent weight of crystalline oxalic acid is

A
30

B
63

C
53

D
45

Solution

Equivalent weight =\(\frac{Molecular\,weight}{ Valency}\)

Molecular weight of

\( OOH-C-OOH .2H_2O\) = \(126\over2\)

= 63
Question 8
1 / -0

When potassium permanganate is titrated against ferrous ammonium sulphate, the equivalent weight of potassium permanganate is

A
\(\frac{Molecular\,weight}{10}\)

B
\(\frac{Molecular\,weight}{5}\)

C
\(\frac{Molecular\,weight}{2}\)

D
Molecular weight

Solution

\(E =\frac{M}{5}\)
Question 9
1 / -0

The specific heat of a metal is 0.16 its approximate atomic weight would be

A
32

B
16

C
40

D
64

Solution

\(Sp. heat \times atomic wt.= 6.4 \)

\(0.16\, \times\,atomic \,wt.= \frac{6.4}{Atomic \,wt.}\)

\( \frac{6.4}{0.16}\)

\(= 40.\)
Question 10
1 / -0

The molecular weight of a gas is 45. Its density at STP is

A
22.4

B
11.2

C
5.7

D
2.0

Solution

The density of gas = molecular wt. of metal/volume

\(=\frac{45}{22.4}\)\(\)

\(2gmlitre^{-1}\)
Question 11
1 / -0

A molar solution is one that contains one mole of a solute in

A
1000 g of the solvent

B
One litre of the solvent

C
One litre of the solution

D
22.4 litres of the solution

Solution

A molar solution is one that contains one mole of a solute in one litre of the solution.
Question 12
1 / -0

One mole of calcium phosphide on reaction with excess of water gives

A
One mole of phosphine

B
Two moles of phosphoric acid

C
Two moles of phosphine

D
One mole of phosphorus pentoxide

Solution

\(Ca_3P_2\,+\,6H_2O\,\to\,2PH_3\,+\,3Ca(OH)_2\)
Question 13
1 / -0

The numbers of moles of \(BaCO_3\) which contain 1.5 moles of oxygen atoms is

A
0.5

B
1

C
2

D
\(6.02\times10^{23}\)

Solution

3 moles of oxygen are that in 1 mole of \(BaCO_3\)

1.5 moles of oxygen are that in mole of \(BaCO_3\)

= \(\frac{1}{3}\times1.5\)

= \(\frac{1}{2}\)

= 0.5
Question 14
1 / -0

2g of oxygen contains number of atoms equal to that in

A
0.5g of hydrogen

B
4g of sulphur

C
7g of nitrogen

D
2.3g of sodium

Solution

Number of atoms in 1 g of an element

= \(6.023\times10^{23} \over at.mass\)

∴ Number of atoms in 2 g of oxygen

= \(6.023\times10^{23} \over 16\)\(\times\,2\) = \(6.023\times10^{23} \over 8\)

Now, number of atoms in 4 g of sulphur

\(6.023\times10^{23} \over 32\)\(\times\,4\) = \(6.023\times10^{23} \over 8\)
Question 15
1 / -0

Irrespective of the source, pure sample, of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of

A
Conservation of mass

B
Multiple proportions

C
Constant composition

D
Constant volume

Solution

The H: O ratios in water are fixed, irrespective of its source. Hence it is law of constant composition.
Question 16
1 / -0

Assuming fully decomposed, the volume of \(CO_2\) released at STP on heating 9.85 g of \(BaCO_3\) (Atomic mass, Ba = 137) will be

A
1.12L

B
2.24L

C
4.06L

D
0.84L

Solution

\(BaCO_3 \,\to\,BaO\,+\,CO_2 \)

192 g of \(BaCO_3\) gives 1 mol of \(CO_2\) = 22.4L

9.85 g of  \(BaCO_3\) will give 0.05 mol of \(CO_2\) which is equal to 1.12 litres.
Question 17
1 / -0

10 mL of 2 M NaOH solution is added to 200 mL of 0.5 M of NaOH solution. What is the final concentration of the solution?

A
0.57 M

B
5.7 M

C
11.4 M

D
1.14 M

Solution

From molarity equation

\(M_1V_1 + M_2V_2\)

\(=\,MV_{total}\,\,2\,\times\) \(10\over 1000\, \) + \(0.5\,\times\,{200\over 1000}\)

= \(M\,\times\,{210\over 1000}\)

\(120\,=\,M\,\times\,210\,M\)

= \(120\over 210\)

= 0.57 M
Question 18
1 / -0

The ppm level of \(F^-\) in a 500 g sample of a tooth paste containing 0.2g \(F^-\) is

A
400

B
1000

C
250

D
200

Solution

\(ppm\,=\,{wt.\,of\,solute\over wt.\,of\,solute}\times10^6\)

\(\,=\,{0.2\over 500}\times10^6\)

= 400
Question 19
1 / -0

Which of the following changes with increase in Temperature?

A
Molality

B
Weight fraction of solute

C
Fraction of solute present in water

D
Mole fraction

Solution

The volume of water increases with temperature.
Question 20
1 / -0

A student performs a titration with different burettes and finds titre values of 25.2 mL, 25.25 ml and 25.0 mL The number of significant figures in the average titre value is

A
3

B
4

C
5

D
6

Solution

A student performs a titration with different burettes and finds titter values of 25.2 mL, 25.25 ml the number of significant figures in the average titter value is 3. This is according to the rules of significant figures.

If two or more numbers having different precision are to be added or subtracted, the answer should be taken as the number of decimal places in the number having less decimal places.

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Some Basic Concepts of Chemistry Test - 13

Result

Some Basic Concepts of Chemistry Test - 13

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SHARING IS CARING

Question 1

Energy

Force

Energy per unit volume

Force per unit volume

Solution

Question 2

5

3

2

4

Solution

Question 3

Oxygen

Hydrogen

Carbon

Iron

Solution

Question 4

5

6

3

2

Solution

Question 5

Metals

Non-metals

Metalloids

None of these

Solution

Question 6

n - m = p − q

n + m =p + q

n = m

p = q

Solution

Question 7

30

63

53

45

Solution

Question 8

\(\frac{Molecular\,weight}{10}\)

\(\frac{Molecular\,weight}{5}\)

\(\frac{Molecular\,weight}{2}\)

Molecular weight

Solution

Question 9

32

16

40

64

Solution

Question 10

22.4

11.2

5.7

2.0

Solution

Question 11

1000 g of the solvent

One litre of the solvent

One litre of the solution

22.4 litres of the solution

Solution

Question 12

One mole of phosphine

Two moles of phosphoric acid

Two moles of phosphine

One mole of phosphorus pentoxide

Solution