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Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

1 g of $^{12}C$ contains $6.022 \times 10^{23}$ atoms of the isotope.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Because according to Avegadro's law,

$1$ mole of $C$ contains $6.022\times { 10 }^{ 23 }$ atoms and $1$ mole of $C$ weighs $12gm$
$\therefore$  $12g$ of $C$ weighs $6.022\times { 10 }^{ 23 }$ atoms

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Calculate the number of iron atoms in a piece of iron weighing $2.8 g$. (Atomic mass of iron $=56$)

  1. $30.11\times { 10 }^{ 23 }$ atoms
  2. $3.11\times { 10 }^{ 23 }$ atoms
  3. $3.0115\times { 10 }^{ 22 }$ atoms
  4. $301.1\times { 10 }^{ 23 }$ atoms
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that, weight of iron =2.8 g
Atomic mass= 56
moles of iron$= \dfrac{\text{wt. of iron}}{\text{atomic mass}}=\dfrac{2.8}{56}=0.05\ moles$
1 mole $= 6.022 \times 10^{23}\ atoms$
0.05 moles $= 6.022 \times 0.05 \times 10^{23}\ atoms=3.0115 \times 10^{22}$ atoms
Option C is correct.
Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

The value of the Avogadro constant is:

  1. $6.022 \times 10^{13}$
  2. $6.022 \times 10^{22}$
  3. $6.022 \times 10^{23}$
  4. $6.022 \times 10^{24}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Avogadro constant $=6.022\times { 10 }^{ 23 }$


Avogadro constant is the number of constituent particles, usually atoms or molecules, that are contained in the amount of the substance given by one mole.

Therefore, the option is C.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

What is the mass of $6.022\times { 10 }^{ 23 }$ molecules of ${NH} _{3}$?

  1. $45.06g$
  2. $19.06\ g$
  3. $17.04g$
  4. $31.02g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$1\space mole$ of $NH _3$ has $6.023 \times 10^{23}$ molecules of $NH _3$.

So, the mass of $6.023 \times 10^{23}$ molecules of $NH _3$. is the mass of $1\space mole$ of $NH _3$  i.e. molar mass of $NH _3.$ 
Molar mass $= 14.01 + 3(1.01) = 17.04 \space g$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

What would be the approximate weight of $1.204\times 10^{24}$ bromine atoms?

  1. 80 grams

  2. 120 grams

  3. 160 grams

  4. 180 grams

  5. 200 grams
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The atomic weight of bromine is 80.

So, 80-gram bromine contains Na atoms.
So, $1.204\times10^{24}$ bromine atoms will weigh:  $80\div 6.02\times10^{23}\times 1.204\times10^{24}=160\ gram$

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

How many molecules of water are present in a $0.25\ mole$ of $H _2O$?

  1. $6.0\times{10}^{22}$
  2. $4.5\times 10^{23}$
  3. $1.5\times{10}^{23}$
  4. $18\times 10^{23}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$1\space mole$ of a compound has $6.023\times 10^{23}$ molecules in it.

$\Rightarrow 1\space mole =$$6.023\times 10^{23}$ molecules
$ 0.25\space mole = x$ molecules
$\Rightarrow x = (\dfrac{6.023}{4}) \times 10^{23}$$= 1.5\times 10^{23}$ molecules of water.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

How many atoms of hydrogen are present in $7.8\ g$ of $Al{(OH)} _{3}$?

  1. $6.0\times {10}^{22}$
  2. $1.8\times {10}^{23}$
  3. $1.7\times {10}^{23}$
  4. $5.1\times {10}^{22}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molar mass of $Al(OH) _3 = 27 + 3\times 16 + 3\times 1 = 78\space g$

So, no. of moles of $Al(OH) _3 = \dfrac{7.8}{78} = 0.1\space moles$
$1\space mole$ of $Al(OH) _3$ has $6.023 \times 10^{23}$ atoms.
So, $0.1\space moles$ has $6.023 \times 10^{22}$ atoms.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

$2K(s)+2{H} _{2}O(l)\rightarrow 2KOH(aq)+{H} _{2}(g)$
If $3.0$ moles of potassium react with excess water, what volume of hydrogen gas will be produced?

  1. $1.5L$
  2. $22.4L$
  3. $67.2L$
  4. $33.6L$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $2K(s) + 2H _2O(l) \rightarrow 2KOH(aq) + H _2 (g)$

$2\space moles$ of Potassium on reaction gives $1 \space mole$ of Hydrogen [ 22.4 at STP]
$2\space moles$ of $K \rightarrow 22.4 \space Litres $ of $H _2$
$2\space moles$ of $K = \dfrac{3\times 22.4}{2} = 33.6 \space Litres$ of $H _2$
So, $33.6\space Litres$ of Hydrogen are produced.

Multiple choice chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

Hydrogen, oxygen and carbon dioxide are taken in containers of $2 l$ volume each. Compare the ratio of the number of molecules of the three gases respectively, under same conditions of temperature and pressure.

  1. 1:8:22

  2. 1:1:1

  3. 1:16:44

  4. 1:8:44
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using ideal gas law: PV =nRT

n = $\dfrac{V}{RT}$
here P, R and T= Constant, V = 2l
hence number of moles also equal  
the ration of moles equal to 1:1:1
 answer is B