Atoms and Molecules Test

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Atoms and Molecules Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

The average relative atomic mass of chlorine is 35.45. It consists of two naturally occurring isotopes chlorine-35 and chlorine-37. What is the fractional abundance of chlorine-37?

A
0.3650

B
0.2200

C
0.3550

D
0.4500

Solution
Question 2
1 / -0

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide is 36. The mixture contains 280 g of nitrogen. Therefore, the amount of $${ CO }_{ 2 }$$ present in the mixture is :

A
440 g

B
44 g

C
4.4 g

D
880 g

Solution

Moles of $$N_2 = \dfrac { 280 }{ 28 } = 10\quad moles$$
Assume, moles of $$CO_2=x$$.
So, average molecular mass of mixture $$ = \dfrac { 10\times28+x\times44 }{ 10+x } $$ $$=36$$ (given)
So, $$ x=10$$ moles
Therefore, weight of $$CO_2 = 10\times 44 = 440$$ g
Question 3
1 / -0

Which among the following is true about one mole of a gas?

A
It always occupies 1 litre.

B
It always occupies 2 litres.

C
It can occupy any volume at S.T.P.

D
It always occupies a fixed volume at N.T.P.
Question 4
1 / -0

$$2.40\ g$$ of the chloroplatinate of a mono acid-base on ignition gave $$0.8\ g$$ of platinum. Calculate the molecular weight of the base. (Given the atomic weight of $$Pt = 195\ g/mol$$).

A
$$87.5\ g/mol$$

B
$$34.6\ g/mol$$

C
$$102.4\ g/mol$$

D
$$111.0\ g/mol$$

Solution

The formula of the chloroplatinate is $$(BH)_2[PtCl_6]$$.
$$2.40$$ g of chlorplatinate corresponds to $$0.8$$ g of Platinum.

$$\therefore 195$$ g of platinum will correspond to $$\dfrac {2.40} {0.8} \times 195 = 585\ g$$ of chloroplatinate.

The molecular weight of the base is $$\dfrac {585-409.81} {2} = 87.5\ g/mol$$.
Here $$409.81$$ is the mass of $$PtCl_6$$.
Hence, option $$A$$ is correct.
Question 5
1 / -0

$$0.3$$ g of chloroplatinic acid of an organic diacidic base left $$0.09$$ g of platinum on ignition. The molecular weight of the organic base is :

A
120

B
240

C
180

D
60

Solution

Let, the formula of the chloroplatinate is $$(BH)_2[PtCl_6]$$.
$$0.3$$ g of chlorplatinate corresponds to $$0.09$$ g of Platinum.
$$195$$ g of platinum will correspond to $$\dfrac {0.3} {0.09} \times 195 = 650\ g$$ of chloroplatinate.
The molar mass of $$(BH_2)[PtCl_6]$$ is $$650$$ g/mol.
The molar mass of $$[PtCl_6]$$ is $$409.81$$ g/mol.
Hence, the molar mass of organic base will be $$(650-409.81) =240$$ g/mol.
Question 6
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A glass vessel weighs $$50$$ g when empty, $$148.0$$ g when completely filled with liquid of density $$0.98 g\ { ml }^{ -1 }$$ and $$50.5$$ g when filled with an ideal gas at $$760$$ mm of Hg at $$300$$ K. Determine the molecular weight of the gas.

A
183 g/mol

B
143 g/mol

C
123 g/mol

D
150 g/mol

Solution

Volume of container $$=\dfrac {mass\ of\ liquid} {density} = \dfrac {148.0-50} {0.98}=100\ ml = 0.1 \ L$$.

Mass of gas $$=50.5 - 50 =0.5 g$$.

The molecular weight can be calculated from the ideal gas equation which is,

$$PV=\frac {W} {M}\times RT$$

$$1\ atm\times 0.1 L=\dfrac {0.5 g} {M}\times 0.082\ L\ atm\ mol^{-1}\ K^{-1} \times 300 K$$

Thus, the molecular weight of the gas is $$123$$ g/mol.
Question 7
1 / -0

What is the molar mass of diacidic organic Lewis base, if $$12$$ g of chloroplatinate salt on ignition produced $$5$$ g residue?

A
$$52$$

B
$$58$$

C
$$88$$

D
None of the above

Solution

The formula of the chloroplatinate salt is $$(BH_2)[PtCl_6]$$.

$$12$$ g of chloroplatinate salt produces $$5$$ g of residue which is platinum.

$$195$$ g of platinum will correspond to $$\dfrac {12}{5} \times 195 = 468$$ g of chloroplatinate salt.

The molar mass of $$(BH_2)[PtCl_6] $$ is $$468$$ g/mol.

The molar mass of $$[PtCl_6]$$ is $$409.81$$ g/mol.

$$ H_2PtCl_6 \rightarrow Pt $$ ;
$$ \dfrac{12}{M_B + 410} = \dfrac{5}{195} = $$ moles of $$Pt$$
Molecular mass of base $$ = 58 $$

Hence, the molar mass of the base $$(BH_2)$$ will be $$58$$ g/mol.
Question 8
1 / -0

The percentage by mole of $${ NO }_{ 2 }(g)$$ in a mixture of $${ NO }_{ 2 }(g)$$ and $$NO(g)$$ having average molecular mass 34 is :

A
25%

B
20%

C
40%

D
75%

Solution

Assume mole % of $$ NO_2 = x $$
Then, mole % of $$ NO = 100-x $$
So, avg. atomic mass of mixture $$ = \dfrac { x\times 46+(100-x)\times 30 }{ 100 } $$ $$= 34$$ (given)
Hence, $$x = 25 $$ %
Question 9
1 / -0

Which of the following statements is correct about the reaction given below?
$$\displaystyle 4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( g \right )$$

A
The total mass of reactants $$=$$ Total mass of the products. It follows the law of conservation of mass.

B
The total mass of reactants $$=$$ Total mass of the products. Therefore, the law of multiple proportions is followed.

C
Amount of $$Fe_{2}O_{3}$$ can be increased by taking any one of the reactants (iron or oxygen) in excess.

D
Amount of $$Fe_{2}O_{3}$$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Solution

Total mass of iron and oxygen in reactants $$=4 (55.85) + 3(32) =319.4\ g$$.

Total mass of iron and oxygen in product $$=2(2 \times 55.85 + 3 \times 8) =319.4\ g$$.

The total mass of iron and oxygen in reactants = the total mass of iron and oxygen in the product, therefore, it follows the law of conservation of mass.

Hence, option $$A$$ is correct.
Question 10
1 / -0

Which of the following is a suitable example for illustrating the law of conservation of mass?
(Atomic mass of O = 16 g/mol, H = 1 g/mole)

A
$$18$$ g of water is formed by the combination of $$16$$ g oxygen with $$2$$ g of hydrogen.

B
$$18$$ g of water in liquid state is obtained by heating $$18$$ g of ice.

C
$$18$$ g of water is completely converted into vapour state on heating.

D
$$18$$ g of water freezes at $$4^oC$$ to give same mass of ice.

Solution

The option (A) is a suitable example for illustrating the law of conservation of mass.

18g of water is formed by the combination of 16g oxygen with 2g of hydrogen.

Total mass of reactants (O and H) = 16 + 2 = 18 g.
The mass of product (water) = 18 g.

Since, the mass of product is equal to the total mass of reactants, the mass is conserved during the reaction, thus illustrating the law of conservation of mass.

Note: The option (A) represents chemical change while all other options represent physical change. The law of conservation of mass corresponds to a chemical change.

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Atoms and Molecules Test - 68

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Atoms and Molecules Test - 68

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SHARING IS CARING

Question 1

0.3650

0.2200

0.3550

0.4500

Solution

Question 2

440 g

44 g

4.4 g

880 g

Solution

Question 3

It always occupies 1 litre.

It always occupies 2 litres.

It can occupy any volume at S.T.P.

It always occupies a fixed volume at N.T.P.

Question 4

$$87.5\ g/mol$$

$$34.6\ g/mol$$

$$102.4\ g/mol$$

$$111.0\ g/mol$$

Solution

Question 5

120

240

180

60

Solution

Question 6

183 g/mol

143 g/mol

123 g/mol

150 g/mol

Solution

Question 7

$$52$$

$$58$$

$$88$$

None of the above

Solution

Question 8

25%

20%

40%

75%

Solution

Question 9

The total mass of reactants $$=$$ Total mass of the products. It follows the law of conservation of mass.

The total mass of reactants $$=$$ Total mass of the products. Therefore, the law of multiple proportions is followed.

Amount of $$Fe_{2}O_{3}$$ can be increased by taking any one of the reactants (iron or oxygen) in excess.

Amount of $$Fe_{2}O_{3}$$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Solution

Question 10

$$18$$ g of water is formed by the combination of $$16$$ g oxygen with $$2$$ g of hydrogen.

$$18$$ g of water in liquid state is obtained by heating $$18$$ g of ice.

$$18$$ g of water is completely converted into vapour state on heating.

$$18$$ g of water freezes at $$4^oC$$ to give same mass of ice.

Solution

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