Tag: gas laws and mole concept

Questions Related to gas laws and mole concept

If the yield of chloroform obtainable from acetone and bleaching powder is $75\%$, what mass of acetone is require for producing $30\ g$ of chloroform? 

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $40\ g$

  2. $9.4\ g$

  3. $10.92\ g$

  4. $14.56\ g$

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Correct Option: A

1 mole of a compound contains 1 mole of C and 2 moles of O. The molecular weight of the compound is:

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. 3

  2. 12

  3. 32

  4. 44

Show answer
Correct Option: D
Explanation:

As 1 mole of a compound contains 1 mole of C and 2 moles of O. 


$ CO _2 \rightarrow12 + 2\times 16= 12 +32 \rightarrow 44 g $

Option D is correct.

What is the weight of $3$ gram atoms of sulphur?

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $96 g$

  2. $99 g$

  3. $100 g$

  4. $3 g$

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Correct Option: A
Explanation:

 The atomic mass of sulphur is 32 g / g atom. The weight of 3 gram atoms of sulphur is $3 \text { g atom } \times 32 \text { g / g atom } =96$ g.

Iron pyrites has formula $FeS _2. (Fe = 56; S = 32)$. What is the mass of sulfur contained in 30 grams of pyrites?

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. 16 g

  2. 32 g

  3. 20 g

  4. 24 g

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Correct Option: A
Explanation:

Molecular weight of $ \displaystyle FeS _2 =  56+2(32)=120$ g/mol.

One mole (120 g) of $ \displaystyle FeS _2$ will contain $ \displaystyle 2 \times 32 = 64$ g of $S$
30 g of $ \displaystyle FeS _2$ will contain $ \displaystyle \dfrac {30}{120} \times  64=16$ g of $S$.
Hence, the mass of sulfur contained in 30 grams of pyrites is 16 g.

An organic compound contains 69% carbon, 4.8% hydrogen and the remaining is oxygen. calculate the masses of carbon dioxide produced when 0.20g of substance is subjected to combustion.
introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. 0.40 g

  2. 0.50 g

  3. 0.60 g

  4. 0.70 g

Show answer
Correct Option: B
Explanation:
Percentage of carbon in organic compound $= 69 \%$

So, 100 g of organic compound contains 69 g of carbon.

∴ 0.2 g of organic compound will contain $=\dfrac{ 69 \times 0.2}{ 100} = 0.138$ g of carbon

The molecular mass of carbon dioxide, $CO _2 = 44$ g

So, 12 g of carbon is contained in 44 g of $CO _2$.

0.138 g of carbon will be contained $= \dfrac{44 \times 0.138}{12} = 0.506$ g of carbon.

Thus, 0.506 g of $CO _2$ will be produced on complete combustion of 0.2 g of an organic compound.

The mass of one molecule of water is approximately:

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $1\ g$

  2. $0.5\ g$

  3. $1.66 \times 10^{-24}\, g $

  4. $ 3 \times 10^{-23}\, g $

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Correct Option: D
Explanation:
Mass of $6.0 \times 10^{23}$ molecules of water is 18 g.

Mass of 1 molecule of water $=\dfrac{M}{N _A}=\dfrac{18}{6.0\times 10^{23}}$

Mass of 1 molecule of water $=3. \times$ $10^{-23}$ g 

In an experiment, the following four gases were produced. 11.2 L of which two gases at STP will weigh 14 g ?

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $N _2O$

  2. $NO _2$

  3. $N _2$

  4. $CO$

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Correct Option: C,D
Explanation:
22.4 L of  a gas at STP$=$ 1 mole.
11.2 L of  a gas at STP$=$ 0.5 mole.
11.2 L of a gas at STP will weigh 14 g.
0.5 moles of a gas at STP will weigh 14 g.
1 mole of a gas at STP will weigh $\dfrac {1}{0.5} \times 14=28$ g.
The molecular weight of the gas is 28 g/mol.
Molecular weight of  $N _2O = 2 (14)+16=44$ g/mol.
Molecular weight of  $NO _2 =14+ 2 (16)=46$ g/mol.
Molecular weight of  $N _2 = 2 (14)=28$ g/mol.
Molecular weight of  $CO = 12+16=28$ g/mol.
Hence, 11.2 L of $N _2$ and $CO$ at STP will weigh 14 g.

If of conservation of mass was to hold true, then 20.8 g of ${ BaCl } _{ 2 }$ on reaction with 9.8 g of ${ H } _{ 2 }{ SO } _{ 4 }$ will produce 7.3 g of $HCl$ and ${ BaSO } _{ 4 }$ equal to :

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. 11.65 g

  2. 23.3 g

  3. 25.5 g

  4. 30.6 g

Show answer
Correct Option: B
Explanation:

$BaCl _2+H _2SO _4 \longrightarrow BaSO _4+2HCl$


$1$ mole of $BaCl _2$ reacts with $1$ mole of $H _2SO _4$ to give $1$ mole of $BaSO _4$ and $2$ moles of $HCl$.


Here, moles of $BaCl _2=\dfrac{20.8}{208}=$ moles of $H _2SO _4= \dfrac{9.8}{98}=0.1$

$\therefore$ Moles of $BaSO _4$ formed $=0.1$

$\therefore$ Mass of $BaSO _4$ formed $=0.1 \times 233= 23.3 g$


i.e. $20.8+9.8=7.3+23.3=30.6$

Hence the correct option is B.

The number of electrons which will together weigh one gram is :

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $1.098 \times 10^{27}$ electrons

  2. $9.1096\times 10^{31}$ electrons

  3. 1 electrons

  4. $1\times 10^4$ electrons

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Correct Option: A
Explanation:

Mass of a electrons = $9.1096\times 10^{-31}Kg$
1g or $10^{-3}kg = \dfrac{1}{9.1096\times 10^{-31}}\times 10^{-3}$
=$1.098\times 10^{27}$ electons

What is the mass of oxalic acid, $ H _{2}C _{2}O _{4},$ which can be oxidized to $ CO _{2}$ by 100 ml of $MnO _{4}^{-}$ solution, 10 ml of which is capable of oxidizing 50 ml of $ 1.00 N\ I^{-} $ to $ I _{2}?$

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. 2.25 g

  2. 52.2 g

  3. 25.2 g

  4. 22.5 g

Show answer
Correct Option: D
Explanation:

Balanced chemical reaction,
$2KMnO _{4}+5H _{2}C _{2}O _{4}+3H _{2}SO _{4}\rightarrow 2MnSO _{4}+10CO _{2}+K _{2}SO _{4}+8H _{2}O$

$(KMnO _{4})N _{1}V _{1}= N _{2}V _{2}(I _{2})$

$N _{1}\times 10= 1\times 50$

$N _{1}= 5N$

n-factor for $KMnO _{4}= 7-2=5$

Moles of $KMnO _{4}=\dfrac{5}{5}=1$

2 mole $KMnO _{4}= 5$ mole $H _{2}C _{2}O _{4}$

1 mole $KMnO _{4}= 2.5$ mole $H _{2}C _{2}O _{4}$

In 100 mL or 0.1 L $= 0.1\times 2.5= 0.25$ moles
Mass of $H _{2}C _{2}O _{4}= 0.25\times 90= 22.5g$