Tag: quantitative chemistry

Questions Related to quantitative chemistry

Which of the following expressions is equal to the number of iron ($Fe$) atoms present in $10.0$ g $\displaystyle Fe$ ? (atomic mass of $\displaystyle Fe$ = $55.9$ amu) 

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $ 10\times 55.9\times( 6.022\times { 10 }^{ 23 }) $ atoms

  2. $\dfrac{( 6.022\times { 10 }^{ 23 })}{10}\times 55.9$ atoms

  3. $ 10\times \dfrac {( 6.022\times { 10 }^{ 23 } )}{ 55.9}$ atoms

  4. $\dfrac {55.9}{10} \times ( 6.022\times { 10 }^{ 23 } ) $

  5. $ \dfrac {10}{ ( 55.9\times 6.022\times { 10 }^{ 23 } )} $ atoms

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Correct Option: C
Explanation:

The expression $\displaystyle 10\times \left( 6.022\times { 10 }^{ 23 } \right)/ 55.9$ is equal to the number of iron (Fe) atoms present in 10.0 g Fe.
The atomic mass of Fe is 55.9 g/mol.
The mass of Fe is 10.0 g. Mass is divided with atomic mass to obtain number of moles.
The number of moles of Fe $ =  \dfrac {10.0}{55.9}$ moles.
The number of moles is multiplied with avogadro's number to obtain the number of Fe atoms.
The number of Fe atoms  $ =  \dfrac {10.0}{55.9} \times 6.023 \times 10^{23}$.

14 g of nitrogen contains $3.01 \times 10^{23}$ nitrogen molecules.

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. True

  2. False

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Correct Option: A
Explanation:

1 mole nitrogen contains $6.02\times10^{23}$ nitrogen molecules.

Thus 14 gram nitrogen will contain $14/28.Na=0.5Na=3.01\times10^{23}$ nitrogen molecules. ( Molecular weight of nitrogen is 28)

Mass of $12.044 \times 10^{23}$ atoms of hydrogen is:

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $1 g$

  2. $2 g$

  3. $3 g$

  4. $4 g$

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Correct Option: B
Explanation:
$\dfrac{mass}{molecular\, weight}=\dfrac{No.\, of\, atoms}{N _{A}}$
$mass=\dfrac{12.044\times 10^{23}}{6.022\times 10^{23}}\times 1$ {Hydrogen molecular weight =$ 1 g/mol$}
$Mass = 2g$

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. True

  2. False

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Correct Option: B
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

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  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

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Correct Option: C
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.

The value of the Avogadro constant is:

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $6.022 \times 10^{13}$

  2. $6.022 \times 10^{22}$

  3. $6.022 \times 10^{23}$

  4. $6.022 \times 10^{24}$

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Correct Option: C
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.

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $45.06g$

  2. $19.06\ g$

  3. $17.04g$

  4. $31.02g$

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Correct Option: C
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 $= 14.01 + 3(1.01) = 17.04 \space g$

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. 80 grams

  2. 120 grams

  3. 160 grams

  4. 180 grams

  5. 200 grams

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Correct Option: C
Explanation:

The atomic weight of bromine is 80.

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

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $6.0\times{10}^{22}$

  2. $4.5\times 10^{23}$

  3. $1.5\times{10}^{23}$

  4. $18\times 10^{23}$

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Correct Option: C
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.

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

chemistry quantitative chemistry avogadro hypothesis avogadro's law avogadro law

  1. $6.0\times {10}^{22}$

  2. $1.8\times {10}^{23}$

  3. $1.7\times {10}^{23}$

  4. $5.1\times {10}^{22}$

Show answer
Correct Option: A
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.