MCQ Questions for Class 9 Chemistry Atoms And Molecules Quiz 12

The average relative atomic mass of chlorine is 35.It consists of two naturally occurring isotopes chlorine-35 and chlorine-What is the fractional abundance of chlorine-37?

0%
0.3650
0%
0.2200
0%
0.3550
0%
0.4500

The average molecular mass of a mixture of gas containing nitrogen and carbon dioxide isThe mixture contains 280 g of nitrogen. Therefore, the amount of

${ CO }_{ 2 }$ present in the mixture is :

0%
440 g
0%
44 g
0%
4.4 g
0%
880 g

Explanation

Moles of $N_2 = \dfrac { 280 }{ 28 } = 10\quad moles$

Assume, moles of $CO_2=x$ .

So, average molecular mass of mixture $= \dfrac { 10\times28+x\times44 }{ 10+x }$ $=36$ (given)

So, $x=10$ moles

Therefore, weight of $CO_2 = 10\times 44 = 440$ g

The time in seconds it would take to spend Avogadro's number of rupees at the rate of

$10$ lakh rupees per second is :

0%
$6.023\times 10^{17}$
0%
$6.023\times 10^{23}$
0%
$6.023\times 10^{10}$
0%
$\displaystyle 6.023\times 10^{19}$

Explanation

The time in seconds it would take to spend Avogadro's number of rupees at the rate of 10 lakh rupees per second is

$\dfrac {6.023 \times 10^{23}} {10^6} = 6.023 \times 10^{17}$ seconds.

Which among the following is true about one mole of a gas?

0%
It always occupies 1 litre.
0%
It always occupies 2 litres.
0%
It can occupy any volume at S.T.P.
0%
It always occupies a fixed volume at N.T.P.

$2.40\ g$ of the chloroplatinate of a mono acid-base on ignition gave

$0.8\ g$ of platinum. Calculate the molecular weight of the base. (Given the atomic weight of

$Pt = 195\ g/mol$ ).

0%
$87.5\ g/mol$
0%
$34.6\ g/mol$
0%
$102.4\ g/mol$
0%
$111.0\ g/mol$

Explanation

The formula of the chloroplatinate is

$(BH)_2[PtCl_6]$ .

$2.40$ g of chlorplatinate corresponds to

$0.8$ g of Platinum.

$\therefore 195$ g of platinum will correspond to

$\dfrac {2.40} {0.8} \times 195 = 585\ g$ of chloroplatinate.

The molecular weight of the base is

$\dfrac {585-409.81} {2} = 87.5\ g/mol$ .

Here

$409.81$ is the mass of

$PtCl_6$ .

Hence, option

$A$ is correct.

$0.3$ g of chloroplatinic acid of an organic diacidic base left

$0.09$ g of platinum on ignition. The molecular weight of the organic base is :

0%
120
0%
240
0%
180
0%
60

Explanation

Let, the formula of the chloroplatinate is

$(BH)_2[PtCl_6]$ .

$0.3$ g of chlorplatinate corresponds to

$0.09$ g of Platinum.

$195$ g of platinum will correspond to

$\dfrac {0.3} {0.09} \times 195 = 650\ g$ of chloroplatinate.
The molar mass of

$(BH_2)[PtCl_6]$ is

$650$ g/mol.
The molar mass of

$[PtCl_6]$ is

$409.81$ g/mol.
Hence, the molar mass of organic base will be

$(650-409.81) =240$ g/mol.

What is the molar mass of diacidic organic Lewis base, if

$12$ g of chloroplatinate salt on ignition produced

$5$ g residue?

0%
$52$
0%
$58$
0%
$88$
0%
None of the above

Explanation

The formula of the chloroplatinate salt is

$(BH_2)[PtCl_6]$ .

$12$ g of chloroplatinate salt produces

$5$ g of residue which is platinum.

$195$ g of platinum will correspond to

$\dfrac {12}{5} \times 195 = 468$ g of chloroplatinate salt.

The molar mass of

$(BH_2)[PtCl_6]$ is

$468$ g/mol.

The molar mass of

$[PtCl_6]$ is

$409.81$ g/mol.

$H_2PtCl_6 \rightarrow Pt$ ;

$\dfrac{12}{M_B + 410} = \dfrac{5}{195} =$ moles of

$Pt$
Molecular mass of base

$= 58$

Hence, the molar mass of the base

$(BH_2)$ will be

$58$ g/mol.

The percentage by mole of

${ NO }_{ 2 }(g)$ in a mixture of

${ NO }_{ 2 }(g)$ and

$NO(g)$ having average molecular mass 34 is :

0%
25%
0%
20%
0%
40%
0%
75%

Explanation

Assume mole % of $NO_2 = x$

Then, mole % of $NO = 100-x$

So, avg. atomic mass of mixture $= \dfrac { x\times 46+(100-x)\times 30 }{ 100 }$ $= 34$ (given)

Hence, $x = 25$ %

Which of the following statements is correct about the reaction given below?

$\displaystyle 4Fe\left ( s \right )+3O_{2}\left ( g \right )\rightarrow 2Fe_{2}O_{3}\left ( g \right )$

0%
The total mass of reactants $=$ Total mass of the products. It follows the law of conservation of mass.
0%
The total mass of reactants $=$ Total mass of the products. Therefore, the law of multiple proportions is followed.
0%
Amount of $Fe_{2}O_{3}$ can be increased by taking any one of the reactants (iron or oxygen) in excess.
0%
Amount of $Fe_{2}O_{3}$ produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess.

Explanation

Total mass of iron and oxygen in reactants

$=4 (55.85) + 3(32) =319.4\ g$ .

Total mass of iron and oxygen in product

$=2(2 \times 55.85 + 3 \times 8) =319.4\ g$ .

The total mass of iron and oxygen in reactants = the total mass of iron and oxygen in the product, therefore, it follows the law of conservation of mass.

Hence, option

$A$ is correct.

A glass vessel weighs

$50$ g when empty,

$148.0$ g when completely filled with liquid of density

$0.98 g\ { ml }^{ -1 }$ and

$50.5$ g when filled with an ideal gas at

$760$ mm of Hg at

$300$ K. Determine the molecular weight of the gas.

0%
183 g/mol
0%
143 g/mol
0%
123 g/mol
0%
150 g/mol

Explanation

Volume of container

$=\dfrac {mass\ of\ liquid} {density} = \dfrac {148.0-50} {0.98}=100\ ml = 0.1 \ L$ .

Mass of gas

$=50.5 - 50 =0.5 g$ .

The molecular weight can be calculated from the ideal gas equation which is,

$PV=\frac {W} {M}\times RT$

$1\ atm\times 0.1 L=\dfrac {0.5 g} {M}\times 0.082\ L\ atm\ mol^{-1}\ K^{-1} \times 300 K$

Thus, the molecular weight of the gas is

$123$ g/mol.

Which of the following is a suitable example for illustrating the law of conservation of mass?

(Atomic mass of O = 16 g/mol, H = 1 g/mole)

0%
$18$ g of water is formed by the combination of $16$ g oxygen with $2$ g of hydrogen.
0%
$18$ g of water in liquid state is obtained by heating $18$ g of ice.
0%
$18$ g of water is completely converted into vapour state on heating.
0%
$18$ g of water freezes at $4^oC$ to give same mass of ice.

The most accurate statement with regard to the significance of Avogadro's number,

$6.02 \times {10}^{23}$ .

0%
It is the conversion factor between grams and atomic mass units
0%
It is a universal physical constant just as the speed of particle
0%
It is the number of particles that is required to fill a $1$ -liter container
0%
It is the inverse diameter of an $H$ atom
0%
It is the number of electrons in the universe

Explanation

Avogadro's number (

$6.022 \times 10^{23}$ ) is the conversion factor between the grams and atomic mass units.
Example:
Atomic mass unit of

$O$ -atom

$= 16 u$
Actual mass of

$O$ -atom

$= \dfrac {16}{6.022 \times 10^{23}} = 2.7 \times 10^{-23} g$
As

$6.022\times 10^{23}$ O-atoms have

$16g$ molar mass, which is the same numerical value for atomic mass but a different unit, i.e.,

$amu$ = atomic mass unit (amu).

$\displaystyle n_{1}$ g of substance X reacts with

$\displaystyle n_{2}$ g of substance Y to form

$\displaystyle m_{1}$ g of substance R and

$\displaystyle m_{2}$ g of substance of S. This reaction can be represented as

$X + Y \rightarrow R + S$ . The relation which can be established in the amounts of the reactants and the products will be :

0%
$\displaystyle n_{1}-n_{2}=m_{1}-m_{2}$
0%
$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$
0%
$\displaystyle n_{1}=n_{2}$
0%
$\displaystyle m_{1}=m_{2}$

Explanation

The relation which can be established in the amounts of the reactants and the products will be as follows:

$\displaystyle \text {Total mass of reactants}=\text {Total mass of products}$

$\displaystyle n_{1}+n_{2}=m_{1}+m_{2}$

This is based on the law of conservation of mass.

The correct increasing order of molecular masses is:

0%
$H_{2}O > H_{2}S > CO_{2} > SO_{2}$
0%
$H_{2}O > H_{2}S < CO_{2} > SO_{2}$
0%
$H_{2}O < H_{2}S < CO_{2} < SO_{2}$
0%
$H_{2}O > H_{2}S > CO_{2} < SO_{2}$

Explanation

(i)

$H_2O \Rightarrow 2+16 = 18 u$

(ii)

$H_2S$

$\Rightarrow 2\times 1+1\times 32=2+32=34 u$

(iii)

$CO_2$

$\Rightarrow 1\times 12+2\times 16=12+32=44 u$

(iv)

$SO_2$

$\Rightarrow 1\times 32+2\times 16=32+32=64 u$

Hence,

$H_2O<H_2S<CO_2<SO_2$ is the correct order of increasing molecular mass.

A charged particle formed by the loss or gain of electrons during a compound formation is called an .....

0%
ion
0%
Cation
0%
Anion
0%
Positron

Molecules are made up of ..........

0%
electrons
0%
protons
0%
neutrons
0%
atoms
0%
none of these

$1.375$ g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was

$1.098$ g. In another experiment,

$1.156$ g of copper was dissolved in nitric acid and the resulting solution was evaporated to dryness. The residue of copper nitrate when strongly heated was converted into

$1.4476$ g of cupric oxide. State the law illustrated by these chemical combinations.

0%
Law of reciprocal proportion
0%
Law of multiple proportion
0%
Law of constant composition
0%
None of these

Explanation

Law of constant composition states that all samples of a given chemical compound have the same elemental composition by mass.

Here, in the first experiment, 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was 1.098 g.

1.375 g of cupric oxide has 1.098 g of copper and

$1.375-1.098 = 0.277$ g of oxygen.

Copper and oxygen are in the ration

$\cfrac{1.098}{0.277} = \cfrac{1}{4}$ .

In the second experiment, 1.156 g of copper was dissolved in nitric acid to form copper nitrate, which on strongly heating got converted into 1.4476 g of cupric oxide.

=1.4476 g of cupric oxide has 1.156 g of copper and

$1.4476-1.156 =0.2911 g$ oxygen.

Copper and oxygen are in the ratio

$\cfrac{1.156}{0.2911} = \cfrac{1}{4}$ .

This means whatever may be the source of a compound, it has the same elemental composition by mass in all samples.

The atomic weight of silicon is

$28.086$ and that of oxygen is

$15.9994$ (on the

$C^{12}$ scale). Calculate the mass in grams of a single molecule of

$SiO_2$ .

0%
$9.977\times 10^{-23}gm$
0%
$9.8\times 10^{-22}gm$
0%
$1\times 10^{-22}gm$
0%
$6.6\times 10^{-23}gm$

Explanation

Molecular weight of

$SiO_{2}$ is

$28.086 + 2\times 15.9994= 60.08$ u

$\therefore$ 1 mole of

$SiO_{2}$ weighs 60.08 g

$\Rightarrow$

$6.022 \times 10^{23}$ moelcules weigh 60.08 g of

$SiO_{2}$ (By Avogadro's Rule)

1 molecule of

$SiO_{2}$ contains

$\dfrac {60.08g}{6.022 \times 10^{23} } \times 1 = 9.977 \times 10^{-23} g$

$\therefore$ Mass of a single molecule of

$SiO_{2} = 9.977 \times 10^{-23}g.$

If 10

$^{21}$ molecules are removed from 200 mg of CO

$_{2}$ , the number of moles of CO

$_{2}$ left is :

0%
$2.88 \times 10^{-3}$
0%
$28.8 \times 10^{-3}$
0%
$0.288 \times 10^{-3}$
0%
$1.66 \times 10^{-3}$

Explanation

Answer:-

Molar mass of

$C{O}_{2}$ = 44g

Given mass = 200 mg = 0.2g

Number of moles =

$\cfrac{0.2}{44} = \cfrac{1}{220}$

$\text{Number of moles} = \cfrac{\text{number of molecules}}{\text{Avogadro's number}}$

$\Rightarrow \text{Number of molecules} = \text{Avogadro's number} \times \text{number of moles}$

$\Rightarrow \text{No. of molecules} = 6.022 \times {10}^{23} \times \cfrac{1}{220} = 2.73 \times {10}^{21} \text{ Molecules in } \cfrac{1}{220} moles$

${10}^{21}$ molecules are removed,

$\therefore$ No. of molecules left =

$2.73 \times {10}^{21} - {10}^{21} = 1.73 \times {10}^{21}$

$\therefore \text{ No. of moles} = \cfrac{\text{No. of molecules}}{\text{Avogadro's number}} = \cfrac{1.73 \times {10}^{21}}{6.23 \times {10}^{23}}$

$= 2.88 \times {10}^{-3}$

Hence, option

$A$ is correct.

Two gaseous samples of the same compound were analyzed. One contained 1.2 g of carbon and 3.2 g of oxygen. The other contained 27.3% carbon and 72.7% of oxygen. The experimental data follows:

0%
Law of conservation of mass
0%
Law of definite proportion
0%
Law of reciprocal proportion
0%
Law of multiple proportion

Explanation

For the first sample, let's calculate the percentages of carbon and oxygen.

Percentage of carbon

$= \dfrac {mass\ of\ carbon}{total\ mass\ of\ substance} \times 100$

$= \dfrac {1.2}{1.2 + 3.2} \times 100$

$= \dfrac {1.2}{4.4} \times 100$

$= 27.27$ %

Percentage of oxygen

$= \dfrac {mass\ of\ oxygen}{total\ mass\ of\ substance} \times 100$

$= \dfrac {3.2}{1.2 + 3.2} \times 100$

$= \dfrac {3.2}{4.4} \times 100$

$= 72.72$ %

For the second sample, the percentages of carbon and oxygen are given:

Percentage of carbon

$= 27.3$ %

Percentage of oxygen

$= 72.7$ %

Both the samples have an equal percentage of the two elements. This follows law of definite proportions: “In a chemical substance the elements are always present in definite proportions by mass”.

If we assume that one-sixth the mass of an atom of

$^{12}C$ isotope is taken as the reference, the mass of one molecule of oxygen will:

0%
be double its original value
0%
be half its original value
0%
be the same
0%
increase by four fold

Explanation

We know that $12\ g\ of\ Carbon=6.023\times \ 10^{23} atoms$

So $1\ g=\frac{6.023\ X\ 10^{23}}{12}$

$6\ g\ of \ carbon=6\times \frac{6.023\ X\ 10^{23}}{12}$

$=\frac{1}{2}\times 6.023\ \times 10^{23}$

This is our new Avogadro's number according to the given conditions.

$1\ atomic\ mass\ unit=\frac{1}{N_A}=\frac{2}{6.023\ X\ 10^{23}}$

Mass of $1$ mole of $O_2=32\times \ Avogrado's\ number\times \frac{2}{6.023\ \times 10^{23}}$

$=32\times \frac{6.023\ \times 10^{23}}{2}\times \frac{2}{6.023\ \times 10^{23}}$

$=32g$

Hence, there is no change in the mass number of $O_2.$

Option $\text{C}$ is correct.

How many oxygen atoms will be present in

$88\ g$ of

$CO_2$ ?

0%
$24.08 \times 10^{23}$
0%
$6.023 \times 10^{23}$
0%
$44 \times 10^{23}$
0%
$22 \times 10^{24}$

Explanation

1 mol of

$CO_2=12+32=44\ g$

Number of oxygen in 1 molecule of

$CO_2=2$

$\therefore$ 44 g

$CO_2$ contains 2 mol oxygen

$88\ g \ CO_2$ contains

$=\cfrac{2}{44} \times 88=4$ mol oxygen

as,

$1 \ mol =6.022 \times 10^{23}$ atoms

$4\ mol\ oxygen =4 \times 6.022 \times 10^{23}$ oxygen atoms

$=24.08 \times 10^{23}$ oxtgen atoms

Hence, option

$A$ is correct.

How many grams of

$CaO$ are required to react with

$852\ g$ of

$P_4O_{10}$ according to below given reaction:

$6CaO+P_4O_{10}\to 2Ca_3(PO_4)_2$

0%
852 g
0%
1008 g
0%
85 g
0%
7095 g

Explanation

The balanced reaction is :

$6CaO + P_4O_{10} \rightarrow 2Ca_3 (PO_4)_2$

Molar mass of

$P_4O_{10}=4 \times 31 + 10 \times 16=284$ g

Moles of

$P_4O_{10} =\frac{mass}{molar\ mass}=\frac{ 852 }{284} = 3$ moles

1 mol of

$P_4O_{10}$ react with 6 mol of

$CaO$

3 moles will react with

$= 3 \times 6 = 18$ mol of

$CaO$

Mass of 1 mol of

$CaO=40+16=56$ g

Mass of 18 moles of

$CaO= 18 \times 56 = 1008$ g

option B is correct.

How many moles of magnesium phosphate,

$Mg_3(PO_4)_2$ will contain 0.25 mole of oxygen atoms ?

0%
$0.02$
0%
$3.125 \times 10^{-2}$
0%
$1.25 \times 10^{-2}$
0%
$2.5 \times 10^{-2}$

Explanation

$1$ molecule of

$Mg_3(PO_4)_2$ has

$8$ atoms of Oxygen.

$\Rightarrow 1$ mole of

$Mg_3(PO_4)_2$ has

$8$ mole of Oxygen atoms.

Now, it is given

$0.25$ mole of oxygen atoms

$\therefore$ Moles of

$Mg_3(PO_4)_2$ having

$0.25$ mole of

$O$ atoms is

$\cfrac {0.25\times 1}{8}$ moles

$=0.03125$ moles

$=3.125\times 10^{-2}$ moles

Hence, option B is correct.

Number of atoms in 558.5 g of Fe:

(Atomic mass of Fe = 55.85 u)

0%
$15.09 \times 10^{25}$
0%
$6.023 \times 10^{24}$
0%
$38$
0%
$5 \times 2\times 6.023 \times 10^{23}$

Explanation

Given mass of Fe =

$558.5\ g$

Molar mass of Fe =

$55.85\ g/mol$ (As we know that atomic mass and molar mass are numerically equal, only units change, and the atomic mass is given in question)

No. of moles of Fe =

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

$\dfrac {558.5}{55.85}$ =

$10\ moles$

Hence, no. of atoms in 10 moles of Fe =

$10 \times Avogadro\ number$ =

$10 \times 6.023 \times 10^{23}$ =

$6.023 \times 10^{24}\ atoms$

Who first gave the concept of 'atom'?

0%
John Dalton
0%
Aristotle
0%
Kanada
0%
Kapila

Specific volume of cylindrical virus particles is

${ 6.02\quad \times }{ 10 }^{ -2 }$ cc/gm whose radius and length are 7

$\mathring A$ and 10

$\mathring A$ , respectively. If

${ N }_{ A }=6.02\times\ { 10 }^{ 23 }$ find molecular weight of virus.

0%
$15.4$ kg/mol
0%
$1.54\quad \times \quad { 10 }^{ 4 }$ kg/mol
0%
$3.08\quad \times \quad { 10 }^{ 4 }$ kg/mol
0%
$1.54\quad \times \quad { 10 }^{ 3 }$ kg/mol

Explanation

Solution:- (A)

$15.4 \; {Kg}/{mol}$

Volume of cylinder

$= \pi {r}^{2}h$

As the virus is cylindrical.

$\therefore$ Volume of one virus

$=$ volume of cylinder

As the radius and length of virus are

$7 \mathring{A}$ and

$10 \mathring{A}$ respectively.

$\therefore$ Volume of

$1$ virus

$= \cfrac{22}{7} \times {7}^{2} \times 10 = 1540 \times {10}^{-24} {cm}^{3} = 1.54 \times {10}^{-21} {cm}^{3}$

$[\because \mathring 1 = 10^{-8} cm^3]$

Volume of

$1$ mole of virus

$= {N}_{A} \times$ volume of

$1$ virus

$\Rightarrow$ Volume of

$1$ mole of virus

$= 6.02 \times {10}^{23} \times 1.54 \times {10}^{-21} = 6.02 \times 1.54 \times {10}^{2} {cm}^{3}$

Given that specific volume of virus is

$6.02 \times {10}^{-2} {{cm}^{3}}/{g}$

$\therefore \text{Molecular weight} = \cfrac{\text{volume of 1 mole}}{\text{Specific volume}} = \cfrac{6.02 \times 1.54 \times {10}^{2}}{6.02 \times {10}^{-2}} = 1.54 \times {10}^{4} \; {g}/{mol} = 15.4 \; {Kg}/{mol}$

Hence, the correct option is A.

How many molecules are present in one gram of hydrogen (

$H_2$ )?

0%
$6.023 \times 10^{23}$
0%
$6.023 \times 10^{22}$
0%
$3.01 \times 10^{23}$
0%
$3.0125 \times 10^{-12}$

Explanation

Atomic mass of

$H$ =

$1\ u$

So, molar mass of

$H$ =

$1\ g/mol$

Molar mass of

$H_2$ =

$2 \times 1$ =

$2\ g/mol$

Given mass of

$H_2$ =

$1\ g$

No. of moles =

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

$\dfrac {1}{2}$ =

$0.5$

No. of molecules =

$Number\ of\ moles \times Avogadro\ number$ =

$0.5 \times 6.022 \times 10^{23}$ =

$3.01 \times 10^{23}$ molecules

If we assume

$\dfrac {1}{6}th$ part instead of

$\dfrac {1}{12}th$ part of weight of

$^{12}C$ as on amu. The molecular mass of water will be

0%
$9$
0%
$18$
0%
$36$
0%
$10.5$

$25\ g$ of

$M{Cl}_{4}$ contains

$0.5$ mole chlorine then its molecular weight is:

0%
$100\ g\ {mol}^{-1}$
0%
$200 \ g\ {mol}^{-1}$
0%
$150\ g\ {mol}^{-1}$
0%
$400\ g\ {mol}^{-1}$

Explanation

From the given weight of

$MCl_4$

$\dfrac{25\ g}{0.5\ mol} \implies \dfrac{50\ g}{1\ mol}$

As there are

$4$ moles of chlorine’s in 1 mole of

$MCl_4$ then molecular weight of

$MCl_4$ can be calculated as,

$\therefore \dfrac{50\ g}{1 \ mol} \times 4\ mol$

$= 200\ g$

Hence

$200\ g\ mol^{ -1 }$ is the answer.

Option

$B$ is correct.

P and Q are two elements which form

${ P }_{ 2 }{ Q }_{ 3 }$ and

${ PQ }_{ 2 }$ . If 0.15 mole of

${ P }_{ 2 }{ Q }_{ 3 }$ weight 15.9 g and 0.15mole of

${ PQ }_{ 2 }$ weight 9.3 g. what are atomic weights of P and Q respectively?

0%
18 and 26
0%
26 and 26
0%
26 and 18
0%
18 and 18

Explanation

Using- no. of mole

$=\cfrac{\text {weight}}{\text{molecular weight}}$

$P_2Q_3 \Rightarrow$

$2P + 3Q =15.9/0.15= 106$ ....(1)

$PQ_2 \Rightarrow$

$P + 2Q = 9.3/0.15=62$ ........(2)

On solving eqn. (1) & (2) we get-

$P =26, Q = 18$

Zinc sulphate contains 22.65%

$Zn$ and 43.9%

${H}_{2}O$ . If the law of constant proportions is true, then the mass of zinc required to give 40 g crystal will be:

0%
9.06 g
0%
90.6 g
0%
0.906 g
0%
906 g

Explanation

Since the percentage of

$Zn$ in the crystal is given as

$22.65$ %, consider that

$100\ g$ of crystals are obtained from

$= 22.65\ g$ of

$Zn$

Applying the law of constant proportions: In a chemical substance the elements are always present in definite proportions by mass.

So, the same proportion of

$Zn$ (

$22.65$ %) will be present in

$40\ g$ crystal as well.

$\therefore 40\ g$ of crystals are obtained from

$=\dfrac { 22.65 }{ 100 } \times40$

$= 9.06\ g\ Zn$

Which of the following cations does not form soluble complex with excess NaOH solution as well as excess

$NH_4OH$ solution ?

0%
$Cu^{+2}$
0%
$Al^{+3}$
0%
$Fe^{+3}$
0%
None of these

Of two oxides of iron, the first contained

$22$ % and the second contained

$30$ % of oxygen by weight. The ratio of weights of iron in the two oxides that combine with the same weight of oxygen is

0%
$1 : 2$
0%
$3 : 2$
0%
$1 : 1$
0%
$2 : 1$

Hydrogen combines with oxygen to form

$H_2O$ in which 16 g of oxygen combine with 2 g of hydrogen. Hydrogen also combines with carbon to form

$CH_4$ in which 2 g of hydrogen combine with 6 g of carbon.If carbon and oxygen combine together then they will do show in the ratio of

0%
13 : 32
0%
6 : 16
0%
1 : 2
0%
12 : 24

The mass of one molecule of carbon dioxide is:

0%
$26.49\times { 10 }^{ 24 }g$
0%
$44\ g$
0%
$7.30\times { 10 }^{ -23 }g$
0%
$22\ g$

Explanation

As we know,

No. of molecules =

$No.\ of\ moles\times Avogadro\ Number$

No. of moles

$= \dfrac{No.\ of\ molecules} {Avogadro\ number}$

Atomic Mass of

$C= 12u$

Atomic Mass of

$O = 16u$

Molecular Mass of

$CO_2=12+(2×16) =44\ u$

Mass of One molecule of

$CO_2= \dfrac{44} {6.023\times 10^{23}}$

$7.30×10^{-23}g$

So, the correct option is

$C$

Molecular mass of one mole of

$H_{2}SO_{4}$ is ______.

0%
$98 a.m.u.$
0%
$1.627\times 10^{-22}g$
0%
Both $A$ & $B$
0%
$1.627\times 10^{-25}g$

Explanation

$Molecular \ mass = mass \ of \ hydrogen \ atom \times number \ of \ atoms + mass \ of \ sulfur \ atom \times number \ of \ atoms + mass \ of \ oxygen \ atom \times number \ of \ atoms$

Molecular mass of

$H_2SO_4 = 1 \times 2 + 32 \times 1 + 16 \times 4 = 98 \ a.m.u.$

Option A is correct.

The weight of one molecule of a compound

$C_{60}$

$H_{122}$ is ?

0%
$1.3 \times 10^{-20}g$
0%
$5.01 \times 10^{-21}g$
0%
$3.72\times 10^{13}g$
0%
$1.4 \times 10^{-21}g$

Atomic radius is measured in?

0%
Centimetres (cm)
0%
Nanometres (nm)
0%
Metres (m)
0%
Kilometres (Km)

If the mass of neutrons is doubled and that of protons is halved, the molecular mass of

$H_2O$ containing only

$H^1$ and

$O^{16}$ atoms will _____________.

0%
increase by about $25\%$
0%
decrease by about $25\%$
0%
increase by about $16\%$
0%
decrease by about $16\%$

$25^oC$ ,

$K_{sp}$ value of

$AgCl$ in water is

$1.8 \times 10^{-10}$ . If

$10^{-5}$ moles of

$Ag^+$ are added to this solution then

$K_{sp}$ will be

0%
$3.13\times 10^{-10}$
0%
$1.8 \times 10^{-10}$
0%
$5.04 \times 10^{-5}$
0%
$1.8 \times 10^{-8}$

The number of moles in

$2 \times 10^{24}$ atoms of iron are

0%
3.3
0%
4.5
0%
5.2
0%
2.1

Explanation

Given, number of atoms of iron

$= 2 \times 10^{24}$

We know, number of moles

$= \dfrac {Given\ number\ of\ particles}{Avogadro\ number}$

$= \dfrac {2 \times 10^{24}} {6.022 \times 10^{23}}$

$=3.3$

Which of the following weighs the least?

0%
2 g atom of N
0%
$3\times 10^{25}$ atoms of carbon
0%
1 mole of sulphur
0%
7 g silver

Number of atoms in 560 g of Fe (atomic mass = 56 u) is

0%
Twice that of 70 g N
0%
Half that of 20 g H
0%
Equal to that of 70 g N
0%
None of these

Explanation

Given, mass of Fe

$= 560\ g$

Atomic mass of Fe

$= 56\ u$

So, molar mass of Fe

$= 56\ gmol^{-1}$

Number of moles

$= \dfrac {Mass}{Molar\ Mass}$

$= \dfrac {560}{56}$

$= 10$

Number of atoms in 560 g Fe

$= Number\ of\ moles \times Avogadro\ number$

$= 10 \times 6.022 \times 10^{23}$

Next, let's find the number of atoms in 70 g N

Given, mass of N

$= 70\ g$

Atomic mass of N

$= 14\ u$

So, molar mass of N

$= 14\ gmol^{-1}$

Number of moles

$= \dfrac {Mass}{Molar\ Mass}$

$= \dfrac {70}{14}$

$= 5$

Number of atoms in 70 g N

$= Number\ of\ moles \times Avogadro\ number$

$= 5 \times 6.022 \times 10^{23}$

Comparing number of atoms in 560 g Fe and 70 g N:

Number of atoms in 560 g Fe

$= 10 \times 6.022 \times 10^{23}$

Number of atoms in 70 g N

$= 5 \times 6.022 \times 10^{23}$

So, number of atoms in 560 g Fe = Twice the number of atoms in 70 g N

The compound which does not exist is

0%
$BF_3$
0%
$AlF_3$
0%
$AlI_3$
0%
$TlI_3$
0%
$TlF_3$

The number of gram molecules of oxygen in

$6.02 \times 10^{24}$ CO molecules is:-

0%
10 $\mathrm{g}$ molecules
0%
5 $\mathrm{g}$ molecules
0%
1 $\mathrm{g}$ molecules
0%
0.5 $\mathrm{g}$ molecules

A gas

$X$ is

$8$ times heavier than hydrogen and the other gas

$Y$ is two times heavier than

$X$ . The mol. wt. of

$Y$ is:

0%
$3$
0%
$8$
0%
$16$
0%
$32$

How many moles of

$CO_{ 2 }$ are present in 220 mg ?

0%
5 moles
0%
0.005 mole
0%
5000 moles
0%
10 moles

Explanation

Weight of given sample of

$CO_2=220mg=220\times 10^{-3}g$

$\text{no. of moles (n) }=\dfrac{\text{Weight of given sample}}{\text{molecular weight}}$

$=\dfrac{220\times 10^{-3}}{44}$

$=5\times 10^{-3}=0.005\,mole$

A gas cylinder of the capacity of 20

$dm^{3}$ is filled with gas

$X$ , the mass of which is

$10\ gm$ . When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is

$2 \ gm$ hence the relative molecular mass of the gas is:

0%
5
0%
10
0%
15
0%
20

Explanation

Correct option is (b)

$\text{Explanation:}$

$\text{Step 1: Finding the vapour density}$

The weight of the

$X$ gas is

$10g$

Since,

$Vapour\ density=\dfrac{Weight\ of\ n\ molecules\ of\ gas}{Weight\ of\ n\ molecules\ of\ hydrogen}$

Therefore,

$Vapour\ density=\dfrac{10}{2}$

$Vapour\ density=5$

$\text{Step 1: Calculation of relative molecular mass}$

Since,

$Relative\ density=2\times vapour\ density$

Therefore,

$Relative\ density=2\times 5$

$Relative\ density=10$

Hence, the relative molecular mass of gas

$X$ is

$10g$ .

Which of the following formulae is not accurately depicted?

0%
Molar mass $=\dfrac {Mass\ of\ substance}{Moles\ of\ substance}$
0%
Number of moles of a substance $=\dfrac {Mass\ of\ substance}{Molar\ mass}$
0%
Number of molecules $=\dfrac {Mass\ of\ the\ substance}{Molar\ mass} \times N_A$
0%
Mole $\times$ Molar mass = Number of molecules

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 12 - MCQExams.com

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Practice Class 9 Chemistry Quiz Questions and Answers

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 12 - MCQExams.com

0:0:1

Practice Class 9 Chemistry Quiz Questions and Answers

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