Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 30

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Question 1
1 / -0

The National Physical Laboratory is situated at:

A
Kolkata

B
New Delhi

C
Bombay

D
None of these

Solution

The CSIR-National Physical Laboratory of India is situated in New Delhi.

It maintains standards of SI units in India and calibrates the national standards of weights and measures.

Hence, option $$B$$ is correct.
Question 2
1 / -0

The number of molecules in 1 L of oxygen and 1 L of nitrogen at STP will be:

A
equal

B
greater for $$N_2$$

C
greater for $$O_2$$

D
equal to Avogadro's number

Solution

Equal volumes of different gases under identical conditions of temperature and pressure contains equal number of molecules. (Avogadro's law)

1 mole = 22.4 litre

Hence, the number of molecules in 1 L of oxygen and 1 L of nitrogen will be equal at STP.
Question 3
1 / -0

When an inflated tyre bursts, the air escaping out will:

A
get heated up

B
be cooled

C
not undergo any change in its temperature

D
be liquified

Solution

$$\text{> Temperature falls when tyre bursts. This is due to adiabatic expension of air.}$$
$$\text{> The pressure of the air inside tyre is sufficiently greater than atmospheric pressure.}$$
$$\text{> When tyre brusts air comes out and work against surrounding.}$$
$$\text{> Due to this internal energy of bursts air decreases and temperature falls to be cooled.}$$
Question 4
1 / -0

Number of water molecules in one litre of pure water is :

A
$$6.023 \times 10^{23}$$

B
$$3.45 \times 10^{25}$$

C
$$55.4 \times 10^{23}$$

D
$$18 \times 10^{23}$$

Solution

One mole of water has a mass of $$18$$ grams and a volume of $$18$$ mL (as the density of water is 1 g per cm$$^3$$).

Therefore, number of moles of water in $$1$$ litre $$= \dfrac{1\times 1000}{18}$$.

Number of molecules $$= $$ Number of moles $$\times N_A$$

$$= \dfrac{1\times 1000}{18} \times 6.023\times 10^{23}$$

$$= 3.45\times 10^{25}$$ molecules.

Hence, the correct answer is option $$\text{B}$$.
Question 5
1 / -0

Photo-electric cell is not used in:

A
television

B
photography

C
reproduction of sound in cinema

D
automatic switching of street lightening circuits

Solution

$$\text{The exposure meter instead of photo-electric cell is used in camera to know the correct time of exposure}$$
Question 6
1 / -0

$$5.6$$ litres of a monoatomic gas at S.T.P contain :

A
$$6.02\times {10}^{23}$$ atoms

B
$$\frac14\times 6.02\times {10}^{23}$$ atoms

C
$$\frac12\times 6.02\times {10}^{23}$$ atoms

D
$$\frac13\times 6.02\times {10}^{23}$$ molecules

Solution

Sicne, $$22.4$$ litres of a gas at S.T.P contains $$6.02\times {10}^{23}$$ atoms.
$$\therefore$$ $$5.6$$ litres of a gas at S.T.P will contain $$\cfrac{5.6}{22.4}\times 6.02\times {10}^{23}=\cfrac{1}{4}\times 6.02\times {10}^{23}$$ atoms.
Question 7
1 / -0

Number of molecules in one litre of water is :

A
$$\displaystyle \frac {6.023}{23.4} \times 10^{23}$$

B
$$18 \times 6.023 \times 10^{23}$$

C
$$\displaystyle \frac {18}{22.4} \times 10^{23}$$

D
$$55.5 \times 6.023 \times 10^{23}$$

Solution

Number of moles of water molecules in 1 litre of water = 55.5 moles
Hence, number of molecules of water = 55.5$$\times$$ Avogadro number
So, option D is correct.
Question 8
1 / -0

In chulhas, gaps are left between the logs:

A
to decrease the ignition temperature of the fuel

B
to allow the air to enter and facilitate fuel burning

C
to cut off the supply of air

D
all of these

Solution

It is important to keep gaps between the logs in chulhas to allow the air to enter and facilitate fuel burning.
Question 9
1 / -0

What is the percentage by weight of sulphuric acid if 13 g of H$$_2$$SO$$_4$$ is dissolved to make 78 g of solution?

A
13.2%

B
14.28%

C
20%

D
16.6%

Solution

Weight of sulphuric acid = 13g
Weight of solution = 78g
Therefore w/W% = $$\frac{13}{78}\times 100$$ = 16.6%
Question 10
1 / -0

Calculate the (w/W)% of 10 g of potassium hydroxide in 40 g of solvent.

A
25%

B
20%

C
30%

D
40%

Solution

Weight of KOH = 10g

Weight of solvent = 40g

Total weight of solution = 40 +10 =50g

Therefore, $$(\dfrac{w}{W})$$% = $$\dfrac{10}{50}\times 100$$ = $$20$$%

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 30

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Some Basic Concepts of Chemistry Test - 30

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SHARING IS CARING

Question 1

Kolkata

New Delhi

Bombay

None of these

Solution

Question 2

equal

greater for $$N_2$$

greater for $$O_2$$

equal to Avogadro's number

Solution

Question 3

get heated up

be cooled

not undergo any change in its temperature

be liquified

Solution

Question 4

$$6.023 \times 10^{23}$$

$$3.45 \times 10^{25}$$

$$55.4 \times 10^{23}$$

$$18 \times 10^{23}$$

Solution

Question 5

television

photography

reproduction of sound in cinema

automatic switching of street lightening circuits

Solution

Question 6

$$6.02\times {10}^{23}$$ atoms

$$\frac14\times 6.02\times {10}^{23}$$ atoms

$$\frac12\times 6.02\times {10}^{23}$$ atoms

$$\frac13\times 6.02\times {10}^{23}$$ molecules

Solution

Question 7

$$\displaystyle \frac {6.023}{23.4} \times 10^{23}$$

$$18 \times 6.023 \times 10^{23}$$

$$\displaystyle \frac {18}{22.4} \times 10^{23}$$

$$55.5 \times 6.023 \times 10^{23}$$

Solution

Question 8

to decrease the ignition temperature of the fuel

to allow the air to enter and facilitate fuel burning

to cut off the supply of air

all of these

Solution

Question 9

13.2%

14.28%

20%

16.6%

Solution

Question 10

25%

20%

30%

40%

Solution

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