Tag: molecular mass

Questions Related to molecular mass

The relative atomic mass of $ Ne$ is $20.$ Its relative molecular mass is:

chemistry basic concepts of chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $10$

  2. $20$

  3. $30$

  4. $40$

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Correct Option: B
Explanation:

The relative atomic mass of  $Ne$ is $20.$ Its relative molecular mass is also $20$ as Neon is mono-atomic.

1 ml ${ O } _{ 2 }$ will be equal to :

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $4g$ equivivalent oxygen

  2. $2g$ equivivalent oxygen

  3. $32g$ equivivalent oxygen

  4. $8g$ equivivalent oxygen

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Correct Option: A
Explanation:

Weight of $1$ mole of $O _2=32g$

$1g$ equivalent is the weight of substance which displaces $8g$ of $O _2$
$\therefore 32g$ of $O _2=\cfrac {32}{8}=4g$ equivalents of oxygen .

How many grams of carbon dioxide is dissolved in 1 litre bottle of carbonated water if the manufacturer uses a pressure of 2.4 atmosphere in the bottling process at 25 degree Celsius?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. 1.52

  2. 4.2

  3. 3.1

  4. 3.52

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Correct Option: D

Common salt obtained from sea-water contains $ 8.775\%$ $NaCl$ by mass. The number of formula units of $NaCl$ present in 25 g of this salt is :

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $3.367 \times { 10 }^{ 23 }$ formula units

  2. $2.258 \times { 10 }^{ 22 }$ formula units

  3. $3.176 \times {10 }^{ 23 }$ formula units

  4. $4.73 \times {10 }^{ 25}$ formula units

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Correct Option: A

How many grams of phosphoric acid $(H _2PO _4)$ would be needed to neutralise $100$g of magnesium hydroxide $(Mg(OH) _2)$?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $66.7$ g

  2. $252$

  3. $112.6$ g

  4. $168$ g

Show answer
Correct Option: C
Explanation:
Apply the formula
${\left( {\dfrac{W}{Equivalent\ wt.}} \right) _{{H _3}P{O _4}}} = \left( {\dfrac{W}{Equivalent\ wt.}} \right) _{Mg{\left( {OH} \right) _2}}$
Hence,
${\dfrac{W}{{98 \times 3}} = \dfrac{{100}}{{58 \times 2}}}$
$\therefore{W = 112.6{\text{ }}gram}$

What is the mass in grams of $6.022\times 10^{23}$ atoms of oxygen?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. 16

  2. 8

  3. 32

  4. 4

Show answer
Correct Option: A
Explanation:
The mass in grams of $6.022\times 10^{23}$ atoms of oxygen is 16 grams. Because $6.022\times 10^{23}$ is avagadro number. Avagadro number of particles are equals to the weight of its molecular weight. So atomic weight of $6.022\times 10^{23}$ atoms of oxygen is  16 which is the atomic weight of oxygen.
Hence option B is correct.

Sulphur trioxide is prepared by the following two reactions:
$S _8(s)+8O _2(g)\rightarrow 8SO _2(g)$
$2SO _2(g)+O _2(g)\rightarrow 2SO _3(g)$
How many grams of $SO _3$ are produced from $1$ mole of $S _8$?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $1280.0$

  2. $640.0$

  3. $960.0$

  4. $320.0$

Show answer
Correct Option: B
Explanation:

${S _8} _{(s)}+{8O _2} _{(g)} \longrightarrow {8SO _2} _{(g)}$

${2SO _2} _{(g)}+{O _2} _{(g)}\longrightarrow {2SO _3} _{(g)}$
By stoichiometry, $1$ mole of ${S _8} _{(s)}$ produces $8$ mole ${SO _2} _{(g)}$
Also, $2$ mole of ${SO _2} _{(g)}$ produces $2$ mole ${SO _3} _{(g)}$
Therefore, $8$ mole ${SO _2} _{(g)}$ produces $8$ moles ${SO _3} _{(g)}$
i.e. $1$ mole ${S _8} _{(s)}$ produces $8$ moles ${SO _3} _{(g)}$
                                         i.e. $8 \times 80g$ of $SO _3$
                                         i.e $640g$ of $SO _3$

Calculate the total volume of $0.1$ molar $KMn{O _4}$ solution that is needed to oxidized 100 mg of each furious oxalate and furious sulphate in a mixture in acidic medium. 

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. 1.096 ml

  2. 1.32 ml

  3. 5.48 ml

  4. None of these

Show answer
Correct Option: C

Consider the following reaction sequence:


${ S } _{ 8 }(s)+{ 80 } _{ 2 }(g)\rightarrow { 8SO } _{ 2 }(g)$

${ 2SO } _{ 8 }(g)+{ O } _{ 2 }(g)\rightarrow { 2SO } _{ 3}(g)$

How many grams of ${ SO } _{ 3 }$ are produced from $1$ mole ${ SO } _{ 8 }$?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $1280 g$

  2. $690 g$

  3. $640 g$

  4. $320 g$

Show answer
Correct Option: C
Explanation:
${{S} _{8}} _{\left( s \right)} + 8 {{O} _{2}} _{\left( g \right)} \longrightarrow 8 {S{O} _{2}} _{\left( g \right)}$

As we have $1$ mole of ${S} _{8}$

According to reaction, one mole of $S{O} _{2}$ produces 8 mles of $S{O} _{2}$

Further,

$2 {S{O} _{8}} _{\left( g \right)} + {{O} _{2}} _{\left( g \right)} \longrightarrow 2 {S{O} _{3}} _{\left( g \right)}$

As per the reaction,

No. of moles of $S{O} _{3}$ produced by $2$ moles of $S{O} _{2} = 2 \text{ moles}$

$\therefore$ No. of moles of $S{O} _{3}$ produced by $8$ moles of $S{O} _{2} = 8 \text{ moles}$

As we know that,

$\text{Wt. of compound} = \text{no. of moles} \times \text{molar mass}$

$\therefore$ Weight of $S{O} _{3}$ in 8 moles $= 8 \times 80 = 640 g \; \left[ \because \text{Molar mass of } S{O} _{3} = 80 g \right]$

Hence, $640$ grams of $S{O} _{3}$ are produced by $1$ mole of ${S} _{8}$.

The correct option is C.

How many grams of $H _{2}SO _{4}$ are present in $0.25\ g$ mole of $H _{2}SO _{4}$?

chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

  1. $2.45$

  2. $24.5$

  3. $0.25$

  4. $245$

Show answer
Correct Option: B
Explanation:
$Moles\quad of\quad { H } _{ 2 }{ SO } _{ 4 }=0.25\\ Molecular\quad mass(M)\quad of\quad { H } _{ 2 }{ SO } _{ 4 }=2+32+16\times 4\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =34+64\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =98\quad amu\\ no.\quad of\quad grams(w)\quad =\quad ?\\ we\quad have,\quad no.\quad of\quad moles\quad =\frac { w }{ M } \\ \qquad \qquad \qquad \qquad \qquad W=\quad no.\quad of\quad moles\times M\\ \qquad \qquad \qquad \qquad \qquad \quad \quad =0.25\times 98\\ \qquad \qquad \qquad \qquad \qquad \quad =24.5g\quad of\quad { H } _{ 2 }{ SO } _{ 4 }$