Atoms and Molecules Test

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Atoms and Molecules Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

The number of gold atoms in 300 mg of a gold ring of 20-carat gold ($$Au = 197$$: pure gold is 24 carat) is:

A
$$4.5 \times {10^{20}}$$

B
$$6.8 \times {10^{15}}$$

C
$$7.6 \times {10^{20}}$$

D
$$9.5 \times {10^{20}}$$

Solution

We have given in the question: $$300mg$$ of a gold ring is $$20\ of\ 24\ Carat$$.
It means in $$300mg$$ ring has $$20$$ part is pure gold and $$4$$ part is other metal.

$$∴$$ Weight of pure gold in $$300\ mg=300\times\ (\frac{20}{24})\times \ 100=250\ mg$$
Now, weight of pure gold $$=250\ mg=0.25g$$
The molecular weight of pure gold $$=197\ g/mol$$
$$∴\ mole=\cfrac{weight\ of\ gold}{molecular\ weight\ of\ gold}$$
$$=\cfrac{0.25}{197}$$

Now,
$$number\ of\ gold\ atoms=number\ of\ mole\ of\ gold\ ×\ 6.023\ ×\ 10^{23}$$

$$=\dfrac{0.25}{197}\ ×\ 6.023\ \times \ 10^{23}$$
$$=7.64\ \times \ 10^{20}$$

Option $$C$$ is correct.
Question 2
1 / -0

In a chemical scale, the relative mass of the isotopic mixture of oxygen atoms $$ (^{16}O,^{17}O,^{18}O) $$ is assumed to be equal to

A
16.002

B
16.00

C
17.00

D
11.00
Question 3
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How many molecules are present in one gram of hydrogen ($$H_2$$)?

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

B
$$6.023 \times 10^{22}$$

C
$$3.01 \times 10^{23}$$

D
$$3.0125 \times 10^{-12}$$

Solution

Atomic mass of $$H$$ = $$1\ u$$
So, molar mass of $$H$$ = $$1\ g/mol$$
Molar mass of $$H_2$$ = $$2 \times 1$$ = $$2\ g/mol$$
Given mass of $$H_2$$ = $$1\ g$$

No. of moles = $$\dfrac {Mass}{Molar\ mass}$$ = $$\dfrac {1}{2}$$ = $$0.5$$

No. of molecules = $$Number\ of\ moles \times Avogadro\ number$$ = $$0.5 \times 6.022 \times 10^{23}$$ = $$3.01 \times 10^{23}$$ molecules
Question 4
1 / -0

If we assume $$\dfrac {1}{6}th$$ part instead of $$\dfrac {1}{12}th$$ part of weight of $$^{12}C$$ as on amu. The molecular mass of water will be

A
$$9$$

B
$$18$$

C
$$36$$

D
$$10.5$$

Solution
Question 5
1 / -0

How many moles of bromine water will be required to complete bromination of 6 moles of phenol?

A
$$6$$

B
$$18$$

C
$$12$$

D
$$3$$

Solution
Question 6
1 / -0

$$25\ g$$ of $$M{Cl}_{4}$$ contains $$0.5$$ mole chlorine then its molecular weight is:

A
$$100\ g\ {mol}^{-1}$$

B
$$200 \ g\ {mol}^{-1}$$

C
$$150\ g\ {mol}^{-1}$$

D
$$400\ g\ {mol}^{-1}$$

Solution

From the given weight of $$MCl_4$$
$$\dfrac{25\ g}{0.5\ mol} \implies \dfrac{50\ g}{1\ mol} $$

As there are $$4$$ moles of chlorine’s in 1 mole of $$MCl_4$$ then molecular weight of $$MCl_4$$ can be calculated as,

$$\therefore \dfrac{50\ g}{1 \ mol} \times 4\ mol$$ $$= 200\ g$$

Hence $$200\ g\ mol^{ -1 }$$ is the answer.
Option $$B$$ is correct.
Question 7
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Two gaseous samples of the same compound were analyzed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% carbon and 72.7% of oxygen. The experimental data follows:

A
Law of conservation of mass

B
Law of definite proportion

C
Law of reciprocal proportion

D
Law of multiple proportion

Solution

For the first sample, let's calculate the percentages of carbon and oxygen.

Percentage of carbon $$= \dfrac {mass\ of\ carbon}{total\ mass\ of\ substance} \times 100$$ $$= \dfrac {1.2}{1.2 + 3.2} \times 100$$ $$= \dfrac {1.2}{4.4} \times 100$$ $$= 27.27$$%

Percentage of oxygen $$= \dfrac {mass\ of\ oxygen}{total\ mass\ of\ substance} \times 100$$ $$= \dfrac {3.2}{1.2 + 3.2} \times 100$$ $$= \dfrac {3.2}{4.4} \times 100$$ $$= 72.72$$%

For the second sample, the percentages of carbon and oxygen are given:
Percentage of carbon $$= 27.3$$%
Percentage of oxygen $$= 72.7$$%

Both the samples have an equal percentage of the two elements. This follows law of definite proportions: “In a chemical substance the elements are always present in definite proportions by mass”.
Question 8
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Zinc sulphate contains 22.65% $$Zn$$ and 43.9% $${H}_{2}O$$. If the law of constant proportions is true, then the mass of zinc required to give 40 g crystal will be:

A
9.06 g

B
90.6 g

C
0.906 g

D
906 g

Solution

Since the percentage of $$Zn$$ in the crystal is given as $$22.65$$%, consider that $$100\ g$$ of crystals are obtained from $$= 22.65\ g$$ of $$Zn$$

Applying the law of constant proportions: In a chemical substance the elements are always present in definite proportions by mass.

So, the same proportion of $$Zn$$ ($$22.65$$%) will be present in $$40\ g$$ crystal as well.

$$\therefore 40\ g$$ of crystals are obtained from $$=\dfrac { 22.65 }{ 100 } \times40$$ $$= 9.06\ g\ Zn$$
Question 9
1 / -0

The weight of one molecule of a compound $$C_{60}$$$$H_{122}$$ is ?

A
$$1.3 \times 10^{-20}g$$

B
$$5.01 \times 10^{-21}g$$

C
$$3.72\times 10^{13}g$$

D
$$1.4 \times 10^{-21}g$$
Question 10
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Of two oxides of iron, the first contained $$22$$% and the second contained $$30$$% of oxygen by weight. The ratio of weights of iron in the two oxides that combine with the same weight of oxygen is

A
$$1 : 2$$

B
$$3 : 2$$

C
$$1 : 1$$

D
$$2 : 1$$

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Atoms and Molecules Test - 71

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Atoms and Molecules Test - 71

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Question 1

$$4.5 \times {10^{20}}$$

$$6.8 \times {10^{15}}$$

$$7.6 \times {10^{20}}$$

$$9.5 \times {10^{20}}$$

Solution

Question 2

16.002

16.00

17.00

11.00

Question 3

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

$$6.023 \times 10^{22}$$

$$3.01 \times 10^{23}$$

$$3.0125 \times 10^{-12}$$

Solution

Question 4

$$9$$

$$18$$

$$36$$

$$10.5$$

Solution

Question 5

$$6$$

$$18$$

$$12$$

$$3$$

Solution

Question 6

$$100\ g\ {mol}^{-1}$$

$$200 \ g\ {mol}^{-1}$$

$$150\ g\ {mol}^{-1}$$

$$400\ g\ {mol}^{-1}$$

Solution

Question 7

Law of conservation of mass

Law of definite proportion

Law of reciprocal proportion

Law of multiple proportion

Solution

Question 8

9.06 g

90.6 g

0.906 g

906 g

Solution

Question 9

$$1.3 \times 10^{-20}g$$

$$5.01 \times 10^{-21}g$$

$$3.72\times 10^{13}g$$

$$1.4 \times 10^{-21}g$$

Question 10

$$1 : 2$$

$$3 : 2$$

$$1 : 1$$

$$2 : 1$$

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