Tag: masses of atoms and molecules

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Questions Related to masses of atoms and molecules

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

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

  1. $66.7$ g
  2. $252$
  3. $112.6$ g
  4. $168$ g
Reveal answer Fill a bubble to check yourself
C Correct answer
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,
${\dfrac{W}{{98 \times 3}} = \dfrac{{100}}{{58 \times 2}}}$
$\therefore{W = 112.6{\text{ }}gram}$
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

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

  1. 16

  2. 8

  3. 32

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

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 $S _8$?

  1. $1280.0$
  2. $640.0$
  3. $960.0$
  4. $320.0$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $SO _3$

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

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. 

  1. 1.096 ml

  2. 1.32 ml

  3. 5.48 ml

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Calculate the milliequivalents of ferrous oxalate and ferrous sulphate. Since KMnO4 acts as an oxidizing agent in acidic medium, use the n-factor to determine the volume of 0.1 M solution required to neutralize the total equivalents.

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

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 ${ SO } _{ 8 }$?

  1. $1280 g$
  2. $690 g$
  3. $640 g$
  4. $320 g$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

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

  1. $2.45$
  2. $24.5$
  3. $0.25$
  4. $245$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Moles\quad of\quad { H } _{ 2 }{ SO } _{ 4 }=0.25\\ Molecular\quad mass(M)\quad of\quad { H } _{ 2 }{ SO } _{ 4 }=2+32+16\times 4\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =34+64\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =98\quad amu\\ no.\quad of\quad grams(w)\quad =\quad ?\\ we\quad have,\quad no.\quad of\quad moles\quad =\frac { w }{ M } \\ \qquad \qquad \qquad \qquad \qquad W=\quad no.\quad of\quad moles\times M\\ \qquad \qquad \qquad \qquad \qquad \quad \quad =0.25\times 98\\ \qquad \qquad \qquad \qquad \qquad \quad =24.5g\quad of\quad { H } _{ 2 }{ SO } _{ 4 }$
Multiple choice chemistry quantitative chemistry molecular mass relative formula mass masses of atoms and molecules

What weight of $SO _2$ can be made by burning sulphur in $5.0$ moles of oxygen?

  1. $640$ grams
  2. $160$ grams
  3. $80$ grams
  4. $320$ grams
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$S+O _2\rightarrow SO _2$


$1$ mole of $O _2$ gives $1$ mole of $SO _2$


Thus $5$ mole of $O _2$ will give 5 mole of $SO _2$

Thus amount of $SO _2$ in 5 mole $SO _2$ is 

$m=5mole\times 64 g/mole\\m=320g$

Thus $5$ mole of $O _2$ will obtain $320g$ of $SO _2$


Hence, the correct option is D.

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

A sample of impure cuprite, $Cu _2O$, contains 66.6% copper. What is the percentage of pure $Cu _2 O$ in the sample: 

  1. 75%

  2. 25%

  3. 60%

  4. 80%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Number of moles of $Cu$ in $66.6g= \cfrac {66.6 g}{63.5}=1.05$ moles

We have $1$ mole of oxygen per $2$ mole of $Cu$

So, moles of oxygen is $0.525$.
Weight of oxygen present= $0.525 mol\times 16g/mol= 8.4g$

So, we have $(66.6+8.4)g=75g$

Our sample is $75$% pure $Cu _2O$.