Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 50

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Question 1
1 / -0

According to Avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain:

A
the same number of molecules

B
different number of molecules

C
the same number of molecules only if their molecular masses are equal

D
the same number of molecules if their densities are equal.

Solution

According to avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain, the same no. of molecules.
For eg. One volume of hydrogen combines with one volume of chlorine to produce two volumes of HCl gas.
$$H_2\\1vol$$ $$+$$ $$Cl_2=\\1 vol$$ $$2HCl\\2 vol$$
Question 2
1 / -0

'Equal volumes of all gases at the same temperature and pressure contain equal number of molecules'. This law is called as

A
Boyle's law

B
Charle's law

C
Avogadro's law

D
Gay Lussac's law

Solution

Avogadro's Law states that 'Equal volumes of all gases at the same temperature and pressure contain equal number of molecules'.
Hence, Option "C" is the correct answer.
Question 3
1 / -0

Thermal electricity is generated from:

A
oil

B
coal

C
gas

D
all of the above

Solution

The electricity that is generated due to heat is thermal electricity. The thermal fuels include, oil, gas, coal, wood-waste, etc. These fuel are used to generate thermal electricity by thermal generation process.
Question 4
1 / -0

A sample of calcium carbonate $$(CaCO_{3})$$ has the following percentage composition: $$Ca = 40\%; \ C = 12\%; \ O = 48\%$$.
If the law of constant proportions is true, then the weight of calcium in $$4 g$$ of a sample of calcium carbonate from another source will be:

A
$$0.016\ g$$

B
$$0.16\ g$$

C
$$1.6\ g$$

D
$$16\ g$$

Solution

In $$100$$ $$g$$ $$CaCO_3$$, $$40$$ $$g$$ $$Ca$$ is present.
In $$4$$ $$g$$ $$CaCO_3$$, Let $$x$$ $$g$$ $$Ca$$ be present
$$\cfrac{x}{4}=\cfrac{40}{100}$$
$$\implies x=1.6$$ $$g$$
Question 5
1 / -0

An impure sample of silver $$(1.5\ g)$$ is heated with $$S$$ to form $$0.124\ g$$ of $$Ag_2S$$. What was the per cent yield of $$Ag_2S$$?

A
$$21.6\%$$

B
$$7.2\%$$

C
$$1.7\%$$

D
$$24.8\%$$

Solution

Reaction involved:
$$2Ag+S \rightarrow Ag_2S$$
216 g pure Ag react with 32 g S to give 248 g $$Ag_2S$$.
To obtain 0.124 g $$Ag_2S$$ amount of Ag reacts with S is calculated as:
$$=\cfrac{216}{248} \times 0.124=0.108$$ g
Thus out of 1.5 g sample only 0.108 g (i.e.pure silver) will react to form $$Ag_2S$$
% yield of $$Ag_2S=\cfrac{0.108}{1.5} \times 100=7.2$$%
Question 6
1 / -0

0.46 g of an organic compound was analysed. The increase in mass of $$ CaCl_2$$ U-Tube was 0.54 g and potash bulb was 0.88 g. The percentage composition of the compound is:

A
C=52.17%, H=13.04%, O=34.79%

B
C=50%,H=50%

C
C=32.19%,H= 18.01%,O=49.8%

D
C=72%,H=28%

Solution

Weight of organic compound = 0.46g

Weight of $${H}_{2}O$$ = 0.54g

Weight of $$C{O}_{2}$$ = 0.88g

$$\% \text{ of H/}{H}_{2}O = \cfrac{2}{18} \times \cfrac{0.54}{0.46} \times 100 = 13.04 \%$$

$$\% \text{ ofC/}C{O}_{2} = \cfrac{12}{44} \times \cfrac{0.88}{0.46} \times 100 = 52.17 \%$$

$$\% \text{ of O} = 100 - (\% \text{ of C} + \% \text{ of H}) = 100 - (52.17 + 13.04) = 34.79\%$$
Question 7
1 / -0

Which scientist discovered that the same amount of space was occupied by equal numbers of molecules of gases, irrespective of whether it was hydrogen or chlorine or fluorine?

A
Gay-Lussac

B
Avogadro

C
Marconi

D
Wohler

Solution

Avogadro first discovered it.
It states that $$1$$ $$mole$$ of any element or compound have $$6.022\times 10^{23}$$ atoms or molecules present at any constant temperature and pressure.
Question 8
1 / -0

Read the passage given below and answer the question:
Chemical reactions involve the interaction of atoms and molecules. A large number of atoms/molecules (approximately $$6.023 \times 10^{23}$$) are present in few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. the following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 faraday =96500 coulombs).
If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is:

A
200

B
225

C
400

D
446

Solution

Moles of $$Na^+$$ ion $$=2$$ $$moles$$
So, maximum $$2$$ $$moles$$ of $$Na-Hg$$ amalgam will be formed.
Weight $$=2\times (23+200)=446$$ $$gm$$
Question 9
1 / -0

The number of atoms in 67.2 L of $${ NH }_{ 3 }$$(g) at STP is:

A
9 $${N }_{ A }$$

B
12 $${N }_{ A }$$

C
3 $${N }_{ A }$$

D
4 $${N }_{ A }$$

Solution

By Avogadro law,
22.4 liter of any gas at STP$$ = $$1mole
$$ =6.022\times { 1 }0^{ 23 }$$ molecules of gas
$$ =$$NA (Avogadro number)

So, in 67.2 L of $${ NH }_{ 3 }=?$$
22.4 L of $$ { NH }_{ 3 }=N_A$$
1 L of $${ NH }_{ 3 }=\cfrac { NA }{ 22.4 } $$

67.2 L of $$ { NH }_{ 3 }=\cfrac { 67.2 }{ 22.4 } \times NA$$
$$ =3N_A$$ molecules

Now, each molecule contain 4 atoma
Number of atoms $$= 4\times 3N_A = 12N_A$$ atoms
Question 10
1 / -0

A $$10.0g$$ a sample of a mixture of calcium chloride and sodium chloride is treated with $$Na_2CO_3$$ to precipitate the calcium as calcium carbonate. This $$CaCO_3$$ is heated to convert all the calcium to $$CaO$$ and the final mass of $$CaO$$ is $$1.62gms$$. The $$\%$$ by mass of $$CaCl_2$$ in the original mixture is :

A
$$15.2\%$$

B
$$32.1\%$$

C
$$21.8\%$$

D
$$11.07\%$$

Solution

$$CaCO_3 +heat \longrightarrow CaO +CO_2$$
$$X$$ $$1.68gm$$
[molecular mass of $$CaCO_3=100u$$
molecular mass of $$CaO=56u$$
cross multiply through the equation and solve for $$x$$ grams of $$CaCO_3$$
$$x=\dfrac{1.62\times 100}{56}$$
$$x =2.89gm$$
$$CaCl_2 +Na_2CO_3 \longrightarrow CaCO_3 +2 NaCl$$
Again cross multiply through the equation and solve for $$x$$ grams of $$CaCl_2$$
$$x=2.89$$

$$x=\dfrac{2.89\times 111}{100}$$
$$x=3.2gm$$
$$\%CaCl_2$$ in original mixture $$3.2grams$$
$$CaCl_2$$ over $$10gms$$ mixture is $$32\%$$

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Some Basic Concepts of Chemistry Test - 50

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Some Basic Concepts of Chemistry Test - 50

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Question 1

the same number of molecules

different number of molecules

the same number of molecules only if their molecular masses are equal

the same number of molecules if their densities are equal.

Solution

Question 2

Boyle's law

Charle's law

Avogadro's law

Gay Lussac's law

Solution

Question 3

oil

coal

gas

all of the above

Solution

Question 4

$$0.016\ g$$

$$0.16\ g$$

$$1.6\ g$$

$$16\ g$$

Solution

Question 5

$$21.6\%$$

$$7.2\%$$

$$1.7\%$$

$$24.8\%$$

Solution

Question 6

C=52.17%, H=13.04%, O=34.79%

C=50%,H=50%

C=32.19%,H= 18.01%,O=49.8%

C=72%,H=28%

Solution

Question 7

Gay-Lussac

Avogadro

Marconi

Wohler

Solution

Question 8

200

225

400

446

Solution

Question 9

9 $${N }_{ A }$$

12 $${N }_{ A }$$

3 $${N }_{ A }$$

4 $${N }_{ A }$$

Solution

Question 10

$$15.2\%$$

$$32.1\%$$

$$21.8\%$$

$$11.07\%$$

Solution

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