Some Basic Concepts Of Chemistry Test 6

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Some Basic Concepts Of Chemistry Test 6

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Weekly Quiz Competition

Question 1
1 / -0

Physical properties are those properties which ______ measured or observed ______ changing the identity or the composition of the substance

A
Cannot be, without

B
Can be, without

C
Cannot be, even by

D
Can be, after

Solution

Physical properties are those which can be measured or observed without changing the identity or composition of the substance.
Question 2
1 / -0

The molar mass of Al₂O₃ is

A
112

B
82

C
102

D
92

Solution

Molar mass = 2(27) +3 (16) = 102g.
Question 3
1 / -0

SI units for Base Physical Quantities of length, mass and current are

A
Meter, kilogram, ampere

B
Meter, gram, ampere

C
Meter, kilogram, candela

D
Decimetre, kilogram, ampere

Solution

M, Kg, A are SI units.
Question 4
1 / -0

Molarity is defined as

A
The number of moles of the solute in 1 m3 of the solution

B
The number of moles of the solvent in 1 litre of the solution

C
The number of moles of the solute in 1 litre of the solution

D
The number of grams of the solute in 1 litre of the solution

Solution

Molarity is defined as no. of moles of solute present per litre of solution.
Question 5
1 / -0

For the reaction
Fe₂ O₃ (s) + 3 CO (g) → 2 Fe (g) + 3 CO₂,
224 g of CO is available to react with 400 g Fe₂O₃, the yield of iron and CO₂, are:

A
225 and 279

B
280 and 330 g

C
210 and 290

D
210 and 279 g

Solution

Moles of CO =8 moles Moles of Fe₂O₃= 2.5 moles.
3 moles of CO is needed for 1 mole of Fe₂O₃ so 8 moles of CO will require 2.66 mole of Fe₂O₃ so Fe₂O₃ is limiting reagent.
1 mole of Fe₂O₃ produce 2 mole of Fe so 2.5 mole of Fe₂O₃will produce 5 mole of Fe = 280g of Fe.
Also 1 mole of Fe₂O₃ also produce 3 mole of CO₂ so 2.5 mole of Fe₂O₃ will produce 7.5 mole of CO₂=330g.
Question 6
1 / -0

According to the law of conservation of mass, a balanced chemical equation has

A
The same number of electrons of each element on both sides of the equation

B
The same number of protons of each element on both sides of the equation

C
The same number of atoms of each element on both sides of the equation

D
The same ratio of atoms of each element on both sides of the equation

Solution

According to this total mass of reactants = total mass of product so no. of atoms of each element in reactant is equal to no. of atoms of that element in product
Question 7
1 / -0

The molar mass of C₆H₁₀O₅ is

A
182

B
152

C
162

D
172

Solution

Molar mass = 6(12) + 10(1) + 5(16) = 162g
Question 8
1 / -0

A measured temperature is 100⁰F on Fahrenheit scale, then what is this reading be on Celsius scale:

A
11.2 ⁰C

B
78 ⁰C

C
102.7 ⁰C

D
37.8 ⁰C

Solution

C-0/100-0 = F-32/180.
C/5= F-32/9.
C/5= 100-32/9.
C/5= 68/9.
C= 68×5/9.
C= 340/9.
C= 37.77.
C= 37.8
Question 9
1 / -0

SI unit of density is

A
kg⁻¹m³

B
kg m⁻²

C
kg m³

D
kg m⁻³

Solution

Density = mass/volume. SI unit of mass is kg and that of volume is m³so SI unit of Density is kg m⁻³
Question 10
1 / -0

The calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction is called

A
Molality

B
Normality

C
Stoichiometry

D
Molarity

Solution

The calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction is called stoichiometry.
Question 11
1 / -0

The molar mass of CaCO₃ is

A
120

B
70

C
100

D
90

Solution

Molar mass = 40 + 12 + 3(16) = 100g
Question 12
1 / -0

There are ____ in 12.0 ml?

A
0.12 L

B
120 L

C
0.012 L

D
12000 L

Solution

Since we know that 1litre -1000ml 1ml-1\1000litre So,12ml-1/1000×12=12/1000 =0.012
Question 13
1 / -0

Molarity of NaOH in a solution prepared by dissolving 4 g of NaOH in enough water to form 250 ml of solution is:

A
0.4 M

B
1.0 M

C
0.1 M

D
4.0 M

Solution
Question 14
1 / -0

The kelvin scale is related to celsius scale by

A
K = °C + 212

B
K = °C + 100

C
K = °C + 32

D
K = °C + 273.15

Solution

K = °C + 273.15
Question 15
1 / -0

Molecular mass of glucose (C₆H₁₂O₆) is

A
180

B
186

C
198

D
192

Solution

Molecular mass will be = 6(12) + 12(1) + 6(16) = 180u.
Question 16
1 / -0

The molar mass of AgNO₃ is

A
169.87 g

B
189.9

C
179.9

D
159.9

Solution

Molar mass of AgNO₃= mass of Ag + N + O₃ = 107.87 + 14 + 3* 16 = 107. 87 + 14 + 48 = 169.87 g
Question 17
1 / -0

The molar mass of ZnSO₄ is

A
161.47 g

B
136.4g

C
166.4g

D
156.4g

Solution

Chemical Formula: ZnSO₄
Molar Mass: 161.47 g/mol (anhydrous)
Question 18
1 / -0

How many atoms of hydrogen are in 67.2 L of H₂ at STP?

A
5.612 × 10²⁴

B
2.612 × 10²⁴

C
4.612 × 10²⁴

D
1.8066 × 10²⁴

Solution

One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
67.2/22.4 = 3 moles of H₂
One mole of any substance contains Avogadro's number of atoms = 6.022 x 10²³ atoms.
So multiply number of moles x number of atoms/mole = 1.8066 x 10²⁴atoms of H₂.
Question 19
1 / -0

In scientific notation for such numbers, any number can be represented in the form N × 10n where

A
n is an exponent having positive or negative values and N can vary from 10 to 100

B
n is an exponent having positive or negative values and N can vary from 0 to 1

C
n is an exponent having positive values only and N can vary from 1 to 10

D
n is an exponent having positive or negative values and N can vary from 1 to 10

Solution

n can vary from to 10 and can be +ve or –ve.
Question 20
1 / -0

Choose the most appropriate answer amongst the options given below for the statement.
A solution of a desired concentration is prepared by diluting

A
Solution of known higher concentration

B
Solution of known lower concentration

C
From a serially diluted solution

D
Stock solution

Solution

Stock solution is diluted to prepare the solution of desired concentration.
Question 21
1 / -0

How many atoms of Oxygen are there in 18g of water?
(Hint: Avogadro’s Number = 6.02 x 10²³ atoms/mol)

A
8.02 x 10²³

B
6.02 x 10²³

C
5 .02 x 10²³

D
7.02 x 10²³

Solution

18g H₂O = 1mol water = 6.02 x 10²³ molecules of water = 6.02 x 10²³ atoms of oxygen.
Question 22
1 / -0

Chemistry does not play a major role in

A
Design and synthesis new materials having specific magnetic, electric and optical properties.

B
Large scale production of a variety of fertilizers

C
Explaining superconductivity

D
Explaining ozone depletion

Solution

Chemistry does not deal in explaining superconductivity.
Question 23
1 / -0

If a matter has definite volume and definite shape, then it is:

A
Solid

B
Liquid

C
Gas

D
All of the Above

Solution

Solid
Holds shape
Fixed Volume

Liquid
Shape of Container
Free Surface
Fixed Volume

Gas
Shape Of Container
Volume of Container

Explanation:
Solid is the only state of matter that has a definite shape and definite volume.
Question 24
1 / -0

The number of moles of solute present in 1 kg of solvent is called:

A
Atomic mass unit

B
Molality

C
Molarity

D
Normality

Solution

Molality is defined as no. of moles of solute present per kg of solvent.
Question 25
1 / -0

There are ____ in 0.05 ml?

A
50 liters

B
0.00005 liters

C
5 liters

D
0.0005 liters

Solution

1L=1000mL so 0.00005L=0.05mL.

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Some Basic Concepts Of Chemistry Test 6

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Question 1

Cannot be, without

Can be, without

Cannot be, even by

Can be, after

Solution

Question 2

112

82

102

92

Solution

Question 3

Meter, kilogram, ampere

Meter, gram, ampere

Meter, kilogram, candela

Decimetre, kilogram, ampere

Solution

Question 4

The number of moles of the solute in 1 m3 of the solution

The number of moles of the solvent in 1 litre of the solution

The number of moles of the solute in 1 litre of the solution

The number of grams of the solute in 1 litre of the solution

Solution

Question 5

225 and 279

280 and 330 g

210 and 290

210 and 279 g

Solution

Question 6

The same number of electrons of each element on both sides of the equation

The same number of protons of each element on both sides of the equation

The same number of atoms of each element on both sides of the equation

The same ratio of atoms of each element on both sides of the equation

Solution

Question 7

182

152

162

172

Solution

Question 8

11.2 0C

78 0C

102.7 0C

37.8 0C

Solution

Question 9

kg−1m3

kg m−2

kg m3

kg m−3

Solution

Question 10

Molality

Normality

Stoichiometry

Molarity

Solution

Question 11

120

70

100

90

Solution

Question 12

0.12 L

120 L

0.012 L

12000 L

Solution

11.2 ⁰C

78 ⁰C

102.7 ⁰C

37.8 ⁰C

kg⁻¹m³

kg m⁻²

kg m³

kg m⁻³

5.612 × 10²⁴

2.612 × 10²⁴

4.612 × 10²⁴

1.8066 × 10²⁴

8.02 x 10²³

6.02 x 10²³

5 .02 x 10²³

7.02 x 10²³