Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

$$10\ g$$ of a crystalline metal sulphate salt when heated generates approximately $$6.4\ g$$ of an anhydrous salt of the same metal. The molecular weight of the anhydrous salt is $$160\ g$$. The number of water molecules present in the crystal is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

Consider that the salt contains $$\times$$ molecules of water.
Molecular weight of anhydrous salt $$= 160 g$$
so molecular weight of hydrated salt will be $$=160+18xg$$
Then, no. of moles of water present in 10x gm of hydrated salt $$=\frac {10}{160+18x}\times x$$
and weight of water present in 10 gm of hydrated salt $$=\frac {10x}{160+18x}\times 18$$
$$\underset {10 g}{Hydrated} salt\rightarrow \underset {604 g}{Anhydrous} salt+\underset {3.6 g}{Water}$$
$$\frac {180x}{160+18x}=3.6$$
$$180x=576+64.8 x$$
$$x=5$$
Question 2
1 / -0

3 g of a hydrocarbon on combustion in excess of oxygen produces 8.8 g of $$CO_2$$ and 5.4 g of $$H_2O$$.

The data illustrates the law of :

A
Conservation of mass

B
Multiple proportions

C
Constant proportions

D
None of the above

Solution

Mass of carbon in 8.8 g $$CO_2 = \dfrac {12}{44} \times 8.8 = 2.4 \ g$$

Mass of hydrogen in 5.4 g $$H_2O = \dfrac{2}{18} \times5.4 = 0.6 \ g$$

Total mass $$( C + H )= 2.4 + 0.6 = 3.0 \ g$$

Mass of hydrocarbon = 3.0 g (given)

These data illustrate the law of conservation of mass.

Option (A) is correct.
Question 3
1 / -0

Temporary hardness is due to bicarbonates of $$Mg^{2+}\, and\, Ca^{2+}$$, it is removed by addition of CaO as follows:
$$Ca(HCO_3)_2\,+\, CaO\, \rightarrow\, 2CaCO_3\,+\, H_2O$$
Mass of $$CaO$$ required to precipitate 2 g $$CaCO_3$$ is:

A
$$2 g$$

B
$$0.56 g$$

C
$$0.28 g$$

D
$$1.12 g$$

Solution

$$Ca(HCO_3)_2 + CaO \rightarrow 2CaCO_3+H_2O$$
Molecular mass of $$CaO=40+16=56 \ g$$
Molecular mass of $$CaO_3=40+12(16 \times 3)=100 \ g$$
Accoridng to the equation, $$2$$ moles of $$CaCO_3$$ i.e $$200 \ g$$ of $$CaCO_3$$ are formed from $$56 \ g \ CaO$$
$$2 \ g$$ of $$CaCO_3$$ will be formed from $$=\cfrac {2 \times 56}{200}=0.56 \ grams$$
Question 4
1 / -0

$$3.01\times 10^{23}$$ molecules of elemental Sulphur will react with 0.5 mole of oxygen gas completely to produce

A
$$6.02\times 10^{23}$$ molecules of $$SO_3$$

B
$$6.02\times 10^{23}$$ molecules of $$SO_2$$

C
$$3.01\times 10^{23}$$ molecules of $$SO_3$$

D
$$3.01\times 10^{23}$$ molecules of $$SO_2$$

Solution

$$\underset {\underset {\frac {1}{2}mole}{1 mole}}{S}+\underset {\underset {\frac {1}{2}mole}{1 mole}}{O_2}\rightarrow \underset {\underset {\frac {1}{2}mole}{1 mole}}{SO_2}$$
$$3.01\times 10^{23} {\;} 0.5 mole {\;} ?$$
$$\therefore 3.01\times 10^{23}$$ molecules of $$SO_2$$ will be formed.
Question 5
1 / -0

A compound was found to contain 5.37% nitrogen. What is the minimum molecular weight of compound ?

A
$$26.07$$

B
$$2.607$$

C
$$260.7$$

D
None of these

Solution

$$5.37 g$$ of $$N$$ in $$100 g$$ compound.

$$14 g N$$= $$\displaystyle\frac{100\times14}{5.37} = 260.7$$ g compound

which is the minimum molecular wt as it should have at least one $$N.$$

Hence (c) is the correct answer.
Question 6
1 / -0

Aluminium reduces manganese dioxide to manganese at high temperature. The amount of aluminium required to reduce one gram mole of manganese dioxide is

A
1/2 gram mole

B
1 gram mole

C
3/4 gram mole

D
4/3 gram mole

Solution

$$4AI + 3MnO_2 \rightarrow 3Mn + 2Al_2 O_3$$
To reduce 3 moles of $$MnO_2$$ required moles of AI = 4
So, for one mole of $$MnO_2$$ required moles of AI will be = 4/3
Question 7
1 / -0

The volume of $$0.5 M$$ aqueous $$NaOH$$ solution required to neutralize $$10 ml$$ of $$2 M$$ aqueous $$HCl$$ solution is

A
$$20 \ ml$$

B
$$40 \ ml$$

C
$$80 \ ml$$

D
$$120 \ ml$$

Solution

$$NaOH {\;} : HCl$$
$$N_1V_1=N_2V_2$$
$$0.5\times V=2\times 10$$
$$V=40 mL$$
Question 8
1 / -0

Four 1-1 litre flasks are separately filled with the gases $$H_2,\, He,\, O_2$$ and ozone. The ratio of total number of atoms of these gases present in different flask would be :

A
1 : 1 : 1 : 1

B
1 : 2 : 2 : 3

C
2 : 1 : 2 : 3

D
3 : 2 : 2 : 1

Solution

According to Avogadro's law, all gases at same temperature and pressure will contain same number of molecules. Hence, number of molecules of all gases will be same bt ration of atoms of each gases will $$2:1:2:3$$.
Question 9
1 / -0

Number of moles of 1 $$m^3$$ gas, at STP, are:

A
44.6

B
40.6

C
42.6

D
48.6

Solution

At STP , volume of 1 mol $$=$$ 22.7 L
1 $$m^3$$ has 1000 L
Then, n (no. of moles) in 1 $$m^3 $$ $$= \displaystyle\frac{1000}{22.7} $$ $$= 44.642 \ mol $$ .
Question 10
1 / -0

Which law stated that 'matter is neither created nor destroyed'?

A
Law of multiple proportion

B
Law of conservation of energy

C
Law of constant composition

D
Law of conservation of mass

Solution

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations.

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

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Some Basic Concepts of Chemistry Test - 28

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Some Basic Concepts of Chemistry Test - 28

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Question 1

$$1$$

$$2$$

$$3$$

$$5$$

Solution

Question 2

Conservation of mass

Multiple proportions

Constant proportions

None of the above

Solution

Question 3

$$2 g$$

$$0.56 g$$

$$0.28 g$$

$$1.12 g$$

Solution

Question 4

$$6.02\times 10^{23}$$ molecules of $$SO_3$$

$$6.02\times 10^{23}$$ molecules of $$SO_2$$

$$3.01\times 10^{23}$$ molecules of $$SO_3$$

$$3.01\times 10^{23}$$ molecules of $$SO_2$$

Solution

Question 5

$$26.07$$

$$2.607$$

$$260.7$$

None of these

Solution

Question 6

1/2 gram mole

1 gram mole

3/4 gram mole

4/3 gram mole

Solution

Question 7

$$20 \ ml$$

$$40 \ ml$$

$$80 \ ml$$

$$120 \ ml$$

Solution

Question 8

1 : 1 : 1 : 1

1 : 2 : 2 : 3

2 : 1 : 2 : 3

3 : 2 : 2 : 1

Solution

Question 9

44.6

40.6

42.6

48.6

Solution

Question 10

Law of multiple proportion

Law of conservation of energy

Law of constant composition

Law of conservation of mass

Solution

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