Tag: quantitative chemistry

Quizzes (10) Questions (111)

Questions Related to quantitative chemistry

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

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) 

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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.