Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 61

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Question 1
1 / -0

What is the correct order of occurance (% by weight) in air of Ne, Ar and Kr ?

A
Ne > Ar > Kr

B
Ar > Ne > Kr

C
Ar > Kr > Ne

D
Ne > Kr > Ar.

Solution
Question 2
1 / -0

$$1$$ mole of oxygen gas weighs ______________

A
$$3g$$

B
$$32g$$

C
$$1g$$

D
$$6.022 \times {10^{22}}$$

Solution
Question 3
1 / -0

What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure, and volume are to be changed from $$220^oC$$,3 atm and 1.65 L to $$110^oC$$, 0.7 atm and 1 L respectively?

A
$$41.4\%$$

B
$$8.18\%$$

C
$$4.14\%$$

D
$$81.8\%$$

Solution
Question 4
1 / -0

Mole fraction of glycerine $$(C_{ 3 }H_{ 5 }(OH)_{ 3 })$$ in a solution of 36 g of water and 46 g of glycerine is :

A
$$0.46$$

B
$$0.36$$

C
$$0.20$$

D
$$0.40$$

Solution

No. of moles of glycerine = 46/92 (where 92 is M.M. of glycerine)
= 0.5moles

No. of moles of water = 36/18 (where 18 is M.M. of water)
= 2moles

So, mole fraction oh glycerine = 0.5/2.5
= 1/5
= 0.20

Hence the correct option is C.
Question 5
1 / -0

0.132 g of an organic compound gave 50 ml of $$N_2$$ at NTP. The weight percentage of nitrogen in the compound is close to

A
15

B
20

C
48.9

D
47.34

Solution
Question 6
1 / -0

The density of gas is 0.0002136 g/mL. Its vapour density is :-

A
12

B
24

C
48

D
32
Question 7
1 / -0

At $${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$$ the density of $${\text{15}}\;{\text{M}}\;{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\;{\text{is}}\;{\text{1}}{\text{.8}}\;{\text{g}}\;{\text{c}}{{\text{m}}^{ - 3}}.$$ Thus, mass percentage of $${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$$ is aqueous solution is

A
2%

B
81.6%

C
18%

D
1.8%

Solution
Question 8
1 / -0

All homologues of which of the following hydrocarbons have the same percentage composition

A
Alkanes

B
Alkynes

C
Alkenes

D
Alkadienes
Question 9
1 / -0

From $$\ 2 \ mg$$ calcium $$1.2\times 10^{19}$$ atoms are removed. The number of $$g$$ - atoms of calcium left is $$(Ca=40)$$:

A
$$5\times 10^{-5}$$

B
$$2\times 10^{-5}$$

C
$$3\times 10^{-5}$$

D
$$5\times 10^{-6}$$

Solution

$$2\ mg$$ Calcium moles $$=\dfrac{2\times 10^{-3}}{40}=5\times 10^{-5}$$

$$1$$ mole $$=6.022\times 10^{23}$$ atoms

$$x=\dfrac{1.2\times 10^{19}}{6.022\times 10^{23}}$$

$$=1.99\times 10^{-5}$$ moles removed

$$\therefore$$ Remaining moles $$=5\times 10^{-5}-1.99\times 10^{-5}$$
$$=3.00\times 10^{-5}$$ moles

$$1\ g$$ atom $$=1$$ mole
$$\therefore$$ Option $$C$$ correct.
Question 10
1 / -0

Molecular mass of dry air is ________________.

A
less than moist air

B
greater than moist air

C
equal to moist air

D
may be greater or less than moist air

Solution

The molecular mass of dry ice is greater than moist air.
As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.

$$M.M $$ of dry air $$\simeq 28.9\ gm$$

$$M.M $$ of moist air $$< 28.9\ gm$$