Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Who performed the gold foil experiment?

A
Thomson

B
Goldstein

C
Chadwick

D
Rutherford

Solution

Rutherford's Gold foil experiment proved the existence of a small massive centre to atoms, which was later known as the nucleus of an atom.

Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold foil experiment to observe the effect of alpha particles on the matter.
Question 2
1 / -0

The weight ratio of roasted ore, coke and limestone fed into the blast furnace in the manufacture of cast iron is:

A
$$8:1:4$$

B
$$6:4:1$$

C
$$8:4:1$$

D
$$8:4:3$$

Solution

The ratio of roasted ore, coke and limestone fed into the blast furnace in the manufacture of cast iron is 8 : 4 : 1.

This mixture is known as charge.

Hence, Option (C) is correct.
Question 3
1 / -0

$$4.6\times 10^{22}$$ atoms of an element weight $$13.8$$ g. What is the atomic mass of the element?

A
290 u

B
180.6 u

C
34.4 u

D
104 u

Solution

$$1$$ mole of any substance contains $$6.02 \times 10^{23}$$ atoms.

Thus, $$4.6 \times 10^{22}$$ atoms corresponds to $$\dfrac{4.6 \times 10^{22}}{6.022 \times 10 ^{23}}=0.0764$$ moles.

$$0.0764$$ moles weighs $$13.8$$ g.

Thus, $$1$$ mole will weigh $$\dfrac{13.8}{0.0764}= 180.6 \ g$$.

Hence, the atomic mass of the element will be 180.6 u.
Question 4
1 / -0

A solution is prepared by adding $$2$$ g of substance $$A$$ to $$18$$ g of water. The mass percent of the solute is:

A
$$10$$

B
$$20$$

C
$$40$$

D
$$25$$

Solution

The formula for the mass percentage of solute in solution is $$=\displaystyle \frac{Mass \ of \ solute}{mass \ of \ solution}\times 100$$
Substituting values in the above expression, we get
Mass percentage $$ =\displaystyle \frac{2}{2+18}\times 100=10 $$%
Question 5
1 / -0

Statement 1: The number of gram molecules of oxygen in $$6.02 \times 10^{24}\ CO$$ molecules is 5.0.
Statement 2: The value of Avogadro number is $$6.02 \times 10^{23}$$

A
Statement 1 is True, statement 2 is True, Statement 2 is a correct explanation of statement 1.

B
Statement 1 is True, statement 2 is True, Statement 2 is not a correct explanation of statement 1.

C
Statement 1 is True, Statement 2 is False.

D
Statement 1 is False, Statement 2 is True.

Solution

$$6.02\times 10^{24}$$ molecules of $$CO=6.02\times 10^{23}\times 10$$ i.e.., 10 moles of CO.
10 moles of CO contains 10 gm atoms of oxygen i.e. 5 gm molecules of oxygen.
hence the correct answer is A the given assertion is correct and reason is the correct explanation to the assertion.
Question 6
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Assertion: 1 mole $$O_{3} (ozone)=N$$ molecule of $$O_{3} (ozone)=3N$$ atoms of $$O (oxygen)=48g$$

Reason: A mole is the amount of matter that contains as many as objects as the number of atoms exactly in 12 g C-12 ($$N$$ = Avogadro number).

A
Assertion is correct but Reason is incorrect.

B
Assertion is incorrect but Reason is correct.

C
Both Assertion and Reason are correct and Reason is correct explanation of Assertion

D
Both Assertion and Reason are correct but Reason is not correct explanation of Assertion

Solution

$$1$$ mole of ozone weighs $$48$$ g (molecular weight). It contains Avogadro number ($$N$$) of molecules. Each ozone molecule contains $$3$$ oxygen atoms. Hence, $$1$$ mole of ozone contains $$3N$$ atoms of oxygen.
Question 7
1 / -0

The mass of hydrogen at STP, that is present in a vessel which can hold $$4$$ grams of oxygen under similar conditions, is:

A
1 gm

B
0.5 gm

C
0.25 gm

D
0.125 gm

Solution

Equal volumes of different gases at STP contains an equal number of moles.

$$4$$ g oxygen corresponds to $$ \cfrac {4} {32} = 0.125$$ mole. This is also equal to the number of moles of hydrogen. $$1$$ mole of hydrogen corresponds to $$2$$ gm.

Hence, $$0.125$$ mole will correspond to $$2 \times 0.125 = 0.25$$ gm.

Hence, the correct option is $$\text{C}$$
Question 8
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Choose the correct option for the following statement.

The use of $$^{12}C$$ scale has superseded the older scale of atomic mass based on $$^{16}O$$ isotope, one important advantage of the former being:

A
The atomic masses on $$^{12}C$$ scale became whole numbers.

B
$$^{12}C$$ is more abundant in the earths crust than $$^{16}O$$.

C
The difference between the physical and chemical atomic masses got narrowed down significantly.

D
$$^{12}C$$ is situated midway between metals and non-metals in the periodic table.

Solution

Physical stuck to pure $$^{16}O$$ for mass scale and chemistry used an average of all three isotopes because natural oxygen is the mixture of $$^{16}O$$, $$^{17}O$$ and $$^{18}O$$. There was not much difference on average.
But with molecules of 200-500 amu, it starts to add up. Use of $$^{12}C$$ which is isotopically pure, get narrow down the difference between physical and chemical atomic masses.
Question 9
1 / -0

Which of the following is the best example to demonstrate the law of conservation of mass?

A
12 gm of carbon combines with 32 gm of oxygen to form 44 gm of $$CO_{2}$$.

B
When 72 gm of carbon is heated in a vacuum and no change in its mass takes place.

C
The weight of a piece of platinum is the same before and after heating in air.

D
None of these

Solution

According to the law of conservation of mass:
The total mass of the reactant = the total mass of the product.

Mass of reactants $$= 12 + 32 = 44$$ i.e w(C)+ w(oxygen)

Mass of products $$= 44$$ i.e. w(carbon dioxide)

Both B and C do not represent a chemical reaction, but only heating of elements, hence they are not the best examples to demonstrate the law of conservation of mass.
Question 10
1 / -0

When 4 L of nitrogen completely reacts with hydrogen, what would be the volume of ammonia gas formed?

A
$$2 L$$

B
$$4 L$$

C
$$5 L$$

D
$$8 L $$

Solution

Reaction:
$$N_2+3H_2 \rightarrow 2NH_3$$
1 lit 3 lit 2 lit

So, 4 litres of nitrogen will produce 8 litres of $$NH_3$$

Option D is correct.

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 20

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Some Basic Concepts of Chemistry Test - 20

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SHARING IS CARING

Question 1

Thomson

Goldstein

Chadwick

Rutherford

Solution

Question 2

$$8:1:4$$

$$6:4:1$$

$$8:4:1$$

$$8:4:3$$

Solution

Question 3

290 u

180.6 u

34.4 u

104 u

Solution

Question 4

$$10$$

$$20$$

$$40$$

$$25$$

Solution

Question 5

Statement 1 is True, statement 2 is True, Statement 2 is a correct explanation of statement 1.

Statement 1 is True, statement 2 is True, Statement 2 is not a correct explanation of statement 1.

Statement 1 is True, Statement 2 is False.

Statement 1 is False, Statement 2 is True.

Solution

Question 6

Assertion is correct but Reason is incorrect.

Assertion is incorrect but Reason is correct.

Both Assertion and Reason are correct and Reason is correct explanation of Assertion

Both Assertion and Reason are correct but Reason is not correct explanation of Assertion

Solution

Question 7

1 gm

0.5 gm

0.25 gm

0.125 gm

Solution

Question 8

The atomic masses on $$^{12}C$$ scale became whole numbers.

$$^{12}C$$ is more abundant in the earths crust than $$^{16}O$$.

The difference between the physical and chemical atomic masses got narrowed down significantly.

$$^{12}C$$ is situated midway between metals and non-metals in the periodic table.

Solution

Question 9

12 gm of carbon combines with 32 gm of oxygen to form 44 gm of $$CO_{2}$$.

When 72 gm of carbon is heated in a vacuum and no change in its mass takes place.

The weight of a piece of platinum is the same before and after heating in air.

None of these

Solution

Question 10

$$2 L$$

$$4 L$$

$$5 L$$

$$8 L $$

Solution

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