Atoms and Molecules Test

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Atoms and Molecules Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following has the highest mass?

A
(a) $$50$$ gms of iron

B
(b) $$5$$ moles of nitrogen gas

C
(c) $$1$$ gm of atom

D
(d)$$ 5 \times 10^{23} $$ atoms of carbon

Solution

Hint: Mass of an atom is directly proportional to number of moles.
The formula used:
Number of moles = $$\dfrac {Mass} {Molar mass}$$
Number of atoms = $$Moles\times6.023\times10^{23}$$
Step 1: The mass of iron in $$50$$ gram.
Here, $$50$$ gram already denotes the mass of iron.
Step 2: The mass of $$5$$ moles of nitrogen gas
According to moles formula mentioned above,
The mass of nitrogen = $$5\times 28=140g$$
Step 3: The mass of $$1$$ gm atoms.
$$1$$ gm atoms= $$1$$ mole of any atom
Step 4: The mass of $$5\times10^{23}$$ atoms of carbon.
Using above formula,
Mass of carbon = $$\dfrac {12\times5\times10^{23}} {6.023\times10^{23}} = 9.96gm$$
Among all , $$5$$ moles of nitrogen has highest mass of $$140g$$
Final answer:
Option $$B$$ is correct answer
Question 2
1 / -0

If the molecular mass of a compound is $$74.5$$ then the compound is:

A
$$LiCl$$

B
$$HCl$$

C
$$NaCl$$

D
$$KCl$$

Solution

$$(1 atom\times 39.0983 Potassium)+(1 atom\times 35.453 Chlorine)$$=$$(74.5513 g/mol )$$, which is the molar mass for $$KCl$$.
Hence, the answer is option $$D$$.
Question 3
1 / -0

A sample of $${ CaCO }_{ 3 }$$ has $$Ca=40$$%, $$C=12$$% and $$O=48$$%. If the law of constant proportion is true then the weight of calcium in $$5g$$ of a sample of $${ CaCO }_{ 3 }$$ from another source will be:

A
$$0.20g$$

B
$$2.0g$$

C
$$2.5g$$

D
$$4.0g$$

Solution

In $$100g$$ of sample $$\rightarrow$$ $$40g$$ of $$Ca$$

$$5g$$ of sample $$\rightarrow$$$$\cfrac { 40 }{ 100 } \times 5=2.0\ g$$ of $$Ca$$
Question 4
1 / -0

Total no. of atoms in $$44\ g$$ of $$CO_{2}$$ is:

A
$$6.02\times 10^{23}$$

B
$$6.02\times 10^{24}$$

C
$$1.806\times 10^{24}$$

D
$$18.06\times 10^{22}$$

Solution

We have given mass of $$CO_2=44\ g$$
Molar mass of $$CO_2=44\ g$$
$$44\ g\ of\ CO_2\ has\ 6.0\times 10^{23}\ molecules\ of\ CO_2$$

(no. of moles $$\dfrac{44}{44}=1$$)

So, No of atoms present in $$44\ g\ of\ CO_2=6.022\times 10^{23}\ atoms$$

1 molecules of $$CO_2=3\ atoms(1C+2O)$$

So, $$44\ g\ of\ CO_2\ has\ 6.022\times 10^{23}\times 3=18.066\times 10^{23}\ atoms$$

So, the answer is $$1.8066\times 10^{24}\ atoms$$
Question 5
1 / -0

In an experiment reproducing the measurements of Rutherford and his co-workers, $$22 \times 10^{-3}$$ mg of He gas was collected in one year from a sample of radium. This sample was observed to emit $$1.06 \times 10^{11} \alpha$$ - particles per second. Thus, Avogadro's number is?

A
$$5.92 \times 10^{23} mol^{-1}$$

B
$$6.92 \times 10^{23} mol^{-1}$$

C
$$6.08 \times 10^{23} mol^{-1}$$

D
$$6.58 \times 10^{23} mol^{-1}$$

Solution

$$ \begin{array}{l} 22 \times 10^{-3} \text {mg per year }=22 \times 10^{-6} \mathrm{~g} \text { peryear } \\ =\frac{22 \times 10^{-6} \mathrm{~g}}{365 \times 24 \times 3600} \\ =6.976 \times 10^{-13} \mathrm{~g} \text { persecond } \\ \text { Thus, mass of } 1.06 \times 10^{11} \alpha \text { particles. } \\ \qquad \begin{aligned} =6.976 \times 10^{-13} \mathrm{~g} \\ \text { } \end{aligned} \end{array} $$
$$4 \mathrm{~g} \mathrm{~mole}^{-1} \quad{ }_{4}\mathrm{He}^{2}(\alpha-$$ particle $$)$$
$$\begin{aligned}=& \frac{1.06 \times 10^{11} \times 4}{6.976 \times 10^{-13}} \\=& 6.08 \times 10^{23}\end{aligned}$$
Correct answer - $$\underline{\underline{C}}$$
Question 6
1 / -0

Find number of oxygen atoms present in $$100 \ mg$$ of $$CaCO_3$$.
(Atomic Mass of $$Ca=40 \ u, \ C=12 \ u, \ O=16 \ $$)

A
$$6.02 \times 10^{23}$$

B
$$6.02 \times 10^{20}$$

C
$$1.806\times 10^{21}$$

D
$$1.204 \times 10^{20}$$

Solution

Moles of $$CaCO_3=\cfrac {100}{10^3 \times 100}=1 \ milli moles$$
Number of oxygen atoms in $$CaCO_3=3$$
Total number of oxygen atoms in $$1 \ milimoles \ CaCO_3=10^{-3} \times 3 \times 6.02 \times 10^{23}=1.806 \times 10^{21}$$
Question 7
1 / -0

A compound contains 3.2% of oxygen. The minimum molecular weight of the compound is:

A
300 u

B
440 u

C
350 u

D
500 u

Solution

The minimum number of oxygen atoms the compound can contain is $$1.$$
Atomic weight of oxygen= $$16 \ u$$
If 3.2 % of the compound weighs $$16 \ u$$
Then 100 % of the compound will weigh = $$\left(\cfrac {100}{3.2}\right)\times 16=500 \ u $$
$$\therefore$$ Minimum molecular weight of the compound = $$500\ u$$
Question 8
1 / -0

Haemoglobin contains $$4.6\%$$ of iron by mass. If the compound contains a single atom of iron then what is its molar mass in $$ { g\ mol }^{ -1 }$$?

A
$$1217.4$$

B
$$1324.7$$

C
$$1232.4$$

D
$$1317.5$$

Solution

The percentage of Iron is $$4.6 \%$$

This means that $$4.6\ g$$ of Iron will be present in $$100 \ g$$ of haemoglobin.

1 g of iron will be present in $$\dfrac{100}{4.6}\ g$$ of haemoglobin.

Molar mass of Iron $$56 \ g \ mol^{-1}$$

$$56\ g \ mol^{-1} $$of iron will be present in $$\dfrac{100\times 56}{4.6} g \ mol^{-1}$$ of haemoglobin.

$$Molar \ mass \ of \ haemoglobin \ is =\dfrac{5600}{4.6}=1217.4\ g\ mol^{-1}$$
Question 9
1 / -0

Which of the following contains the largest mass of hydrogen atoms?

A
$$0.5$$ moles $$C_2H_2O_4$$

B
$$1.1$$ moles $$C_3H_8O_3$$

C
$$1.5$$ moles $$C_6H_8O_6$$

D
$$4.0$$ moles $$C_3H_8O_3$$

Solution

1 mole of $$C_{2}H_{2}O_{4} \longrightarrow$$2 moles of hydrogen=2 g of hydrogen
Thus 0.5 mole of $$C_{2}H_{2}O_{4}$$=1 g of hydrogen

1 mole of $$C_{3}H_{8}O_{3} \longrightarrow$$8 moles of hydrogen=8 g of hydrogen
Thus 1.1 moles of $$C_{3}H_{8}O_{3}$$=8.8 g of hydrogen

1 mole of $$C_{6}H_{8}O_{6} \longrightarrow$$8 moles of hydrogen=8 g of hydrogen
Thus 1.5 moles of $$C_{6}H_{8}O_{6}$$=12 g of hydrogen

1 mole of $$C_{2}H_{4}O_{2} \longrightarrow$$4 moles of hydrogen=4 g of hydrogen
Thus 4 moles of $$C_{3}H_{8}O_{3}$$=16 g of hydrogen

Thus 4 moles of $$C_{3}H_{8}O_{3}$$ contains the largest mass of hydrogen.
Hence, option $$D$$ is correct.
Question 10
1 / -0

A solution of 0.640 g of azulene in 100.0 g of benzene boils at $$80.23^0C$$. The boiling point of benzene is $$80.10^0C$$; the $$K_b$$ is $$ 2.53^0C/molal$$. What is the molecular weight of azulene?

A
108

B
99

C
125

D
134

Solution

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Atoms and Molecules Test - 60

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Atoms and Molecules Test - 60

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Question 1

(a) $$50$$ gms of iron

(b) $$5$$ moles of nitrogen gas

(c) $$1$$ gm of atom

(d)$$ 5 \times 10^{23} $$ atoms of carbon

Solution

Question 2

$$LiCl$$

$$HCl$$

$$NaCl$$

$$KCl$$

Solution

Question 3

$$0.20g$$

$$2.0g$$

$$2.5g$$

$$4.0g$$

Solution

Question 4

$$6.02\times 10^{23}$$

$$6.02\times 10^{24}$$

$$1.806\times 10^{24}$$

$$18.06\times 10^{22}$$

Solution

Question 5

$$5.92 \times 10^{23} mol^{-1}$$

$$6.92 \times 10^{23} mol^{-1}$$

$$6.08 \times 10^{23} mol^{-1}$$

$$6.58 \times 10^{23} mol^{-1}$$

Solution

Question 6

$$6.02 \times 10^{23}$$

$$6.02 \times 10^{20}$$

$$1.806\times 10^{21}$$

$$1.204 \times 10^{20}$$

Solution

Question 7

300 u

440 u

350 u

500 u

Solution

Question 8

$$1217.4$$

$$1324.7$$

$$1232.4$$

$$1317.5$$

Solution

Question 9

$$0.5$$ moles $$C_2H_2O_4$$

$$1.1$$ moles $$C_3H_8O_3$$

$$1.5$$ moles $$C_6H_8O_6$$

$$4.0$$ moles $$C_3H_8O_3$$

Solution

Question 10

108

99

125

134

Solution

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