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Atoms and molecules - class-X

Description: atoms and molecules
Number of Questions: 23
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Tags: atoms and molecules chemistry gas laws and mole concept
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Which of the following molecules specie has the minimum number of lone pairs ? 

  1. $ X _eF _4 $

  2. $ BF^- _4 $

  3. $ ICI _3 $

  4. $ BeF _2 $


Correct Option: D

A quantity of $2.0\ g$ of a triatomic gaseous element was found to occupy a volume of $448\ ml$ at $76\ cm$ of $Hg$ and $273\ K$.

The mass of its each atom is _______________.

  1. $100\ amu$

  2. $5.53\times 10^{-23}\ g$

  3. $33.3\ g$

  4. $5.53\ amu$


Correct Option: B
Explanation:
Given that
$P=76\ cm$ of $Hg =1\ atm$
$V=448\ ml=0.448\ L$
$T=273\ K$

Now,
$PV=nRT$

$n=\dfrac{PV}{RT}=\dfrac{1\times 0.448}{0.082\times 273}=0.02\ moles$

No. of moles $=\dfrac{mass}{molar\ mass}$

Molar mass $=\dfrac{mass}{mole}$

$3\times$ atomic mass $=\dfrac{2}{0.02}$

Atomic mass $=33.3\ g$

$\therefore$ mass of $1\ atom=\dfrac{33.3}{6.022\times 10^{23}}=5.53\times 10^{-23}\ g$

Which of the following gas molecule has the longest mean free path at the same pressure and temperature?

  1. $H _2$

  2. $N _2$

  3. $O _2$

  4. $Cl _2$


Correct Option: A

Two gases X and Y have their molecular speed in ratio of $3:1$ at certain temperature. The ratio of their molecular masses $M _x:M _y$ is?

  1. $1:3$

  2. $3:1$

  3. $1:9$

  4. $9:1$


Correct Option: C

The waste of nuclear power plant contains $C^{12}$ and $C^{14}$ in ratio of $4:1$ by moles. What is the molecular mass of methane gas produced from this disposed of $C^{12}$ and $C^{14}$ are $98\%$ and $2\%$, respectively. 

  1. $15.998$

  2. $16.0053$

  3. $16$

  4. $16.4$


Correct Option: A

A gaseous mixture of three gases A, B and C has a pressure of $10$ atm. The total number of moles of all the gases is $10$. If the partial pressures of A and B are $3.0$ and $1.0$ atm, respectively, and if C has molecular mass of $2.0$, what is the mass of C, in g, present in the mixture?

  1. $6$

  2. $8$

  3. $12$

  4. $3$


Correct Option: A

Which of the following will have the composition (by mass) mass similar as that of acetic acid?

  1. Methyl formate, $HCOOCH _{3}$

  2. Glucose, $C _{6}H _{12}O _{6}$

  3. Formaldehyde, $HCHO$

  4. Formic acid, $HCOOH$


Correct Option: A,B,C

A quantity of $5.0$g of a mixture of He and another gas occupies a volume of $1.5$l at $300$K and $750$mm Hg. The gas freezes at $270$K. At $15$K, the pressure of the gas mixture is $8$mm Hg(at the same volume). What is the molecular mass of the gas?

  1. $103.8$

  2. $4.0$

  3. $495$

  4. $82.24$


Correct Option: A

A protein, isolated from a bovine preparation, was subjected to amino acid analysis. The amino acid present in the smallest amount was lysine, $C _6H _{14}N _2O _2$ and the amount of lysine was found to be $365\ mg$ per $100\ g$ protein. What is the minimum molecular mass of the protein?   

  1. $40,000,000$

  2. $40, 000$

  3. $40$

  4. $4, 00,000$


Correct Option: A

A mixture contains $NaCl$ and unknown chloride $MCl$. When $1\ g$ of this mixture is dissolved in water and excess of $AgNO _{3}$ Solution is added to it, $2.567\ g$ of white precipitate is obtained. In another experiment, $1\ g$ of the same original mixture is heated to $300^{o}C$. Some vapour come out which are absorbed in acidified $AgNO _{3}$ solution by which $1.341\ g$ of white precipitate is formed. The molecular mass of unknown chloride is

  1. $53.4$

  2. $58.5$

  3. $44.5$

  4. $74.4$


Correct Option: A,B

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? 

  1. $40\ g$

  2. $9.4\ g$

  3. $10.92\ g$

  4. $14.56\ g$


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:

  1. 3

  2. 12

  3. 32

  4. 44


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?

  1. $96 g$

  2. $99 g$

  3. $100 g$

  4. $3 g$


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?

  1. 16 g

  2. 32 g

  3. 20 g

  4. 24 g


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.
  1. 0.40 g

  2. 0.50 g

  3. 0.60 g
  4. 0.70 g


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:

  1. $1\ g$

  2. $0.5\ g$

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

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


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 ?

  1. $N _2O$

  2. $NO _2$

  3. $N _2$

  4. $CO$


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 :

  1. 11.65 g

  2. 23.3 g

  3. 25.5 g

  4. 30.6 g


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 :

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

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

  3. 1 electrons

  4. $1\times 10^4$ electrons


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}?$

  1. 2.25 g

  2. 52.2 g

  3. 25.2 g

  4. 22.5 g


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 $H _{2}C _{2}O _{4}= 0.25\times 90= 22.5g$

A definite mass of $ H _{2}O _{2} $ is oxidized by excess of acidified $ KMnO _{4} $ and acidified $ K _{2}Cr _{2}O _{7} $, in separate experiments. Which of the following is/are correct statements? 
(K = 39, Cr = 52, Mn = 55 )

  1. Mass of $ K _{2}Cr _{2}O _{7} $ used up will be greater than that of $ KMnO _{4} $

  2. Moles of $ KMnO _{4} $ used up will be greater than that of $ K _{2}Cr _{2}O _{7} $

  3. Equal mass of oxygen gas is evolved in both the experiments.

  4. If equal volumes of both the solutions are used for complete reaction, then the molarities of $ KMnO _{4} $ and $ K _{2}Cr _{2}O _{7} $ solutions are in $6:5$ ratio.


Correct Option: A,B,D
Explanation:

According to question, reaction of experiment (1) and (2)
(1)$ 5H _{2}O _2+2KMnO _{4}+3H _{2}SO _{4}\rightarrow 5O _{2}+2MnSO _{4}+K _{2}SO _{4}+8H _{2}O $


(2)$ 3H _{2}O _{2}+K _{2}Cr _{2}O _{7}+4H _2SO _{4}\rightarrow Cr _{2}(SO _{4}) _{3}+3O _{2}+K _{2}SO _{4}+7H _{2}O $

(a) According to reaction (1)
5 mole $ H _{2}O _{2} = 2\,mole KMnO _{4} $

1 mole $ H _{2}O _{2} = \dfrac{2}{5} = 0.4\,mole\,KMnO _{4} = 63.2\,g\,KMnO _{4} $

According to reaction (2)
3 mole $ H _{2}O _{2} = 1\,mole\,K _{2}Cr _{2}O _{7} $

1 mole $ H _{2}O _{2} = \dfrac{1}{3} mole\,K _{2}Cr _{2}O _{7} = 98.1 g \, K _{2}Cr _{2}O _{7} $

Mass of $ K _{2}Cr _{2}O _{7}> KMnO _{4} $

(b) Moles of $ KMnO _{4}> $ moles of $ K _{2}Cr _{2}O _{7} (0.333) $

(c)In reaction 1, 5 Mole $ H _{2}O _{2} $ released = 5 mole $ O _{2}\Rightarrow 32\times 5=160g $
In reaction 2, 3 mole $ H _{2}O _{2} $ released = 3 mole $ O _{2}=3\times 32= 96 g$ 

(d) 1 mole of $ H _{2}O _{2} = \dfrac{2}{5} $ moles $ KMnO _{4} $ exp...(1)
1 mole of $ H _{2}O _{2} = \dfrac{1}{3}$ mole $ K _{2}Cr _{2}O _{7} $ exp...(2)
$ \dfrac{KMnO _{4}}{K _{2}Cr _{2}O _{7}} = \dfrac{\dfrac{2}{5}}{\dfrac{1}{3}} = \dfrac{2}{5}\times \dfrac{3}{1} = \dfrac{6}{5}\Rightarrow 6:5 $

Options A, B and D are correct.

Two acids $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed, respectively. Find the ratio of the masses of $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ 

  1. $ 1:1 $

  2. $ 1:2 $

  3. $ 2:1 $

  4. $ 2:3 $


Correct Option: B
Explanation:

$H _{2}SO _{4}+2NaOH\rightarrow Na _{2}SO _{4}+2H _{2}O$

$H _{3}PO _{4}+NaOH\rightarrow NaH _{2}PO _{4}+H _{2}O$

Equivalent of alkali $= 19$ eq of $H _{2}SO _{4}= 1g$ eq of $H _{3}PO _{4}$

Two acids must be reacting in the ratio of their equivalent masses

Eq. wt. of $H _{2}SO _{4}=\dfrac{98}{2}=49$

Eq. wt. of $H _{3}PO _{4}= \dfrac{98}{1}=98$

$\therefore $ ratio of masses of $H _{2}SO _{4}$ & $H _{3}PO _{4}$
$49:98=1:2$

$\Rightarrow 1:2$

A gaseous alkane is exploded with oxygen. The volume of ${O} _{2}$ for complete combustion of alkane to $C{O} _{2}$ formed is in the ratio $7:4$. The molecular formula of alkane is:

  1. ${C} _{2}{H} _{6}$

  2. ${C} _{3}{H} _{8}$

  3. ${C} _{4}{H} _{10}$

  4. $C{H} _{4}$


Correct Option: A
Explanation:

The balanced reaction is given below:


${C} _{n}{H} _{2n+2} +[n+\displaystyle\frac{n+1}{2}]{O} _{2}\rightarrow nC{O} _{2} +(n+1){H} _{2}O$

Given, 

$\displaystyle\dfrac{n+\dfrac{n+1}{2}}{n}=\dfrac{7}{4}\implies n=2$

Hence, the alkane is ${C} _{2}{H} _{6}$.

Hence, the correct option is $A$

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