MCQ Questions for Class 9 Chemistry Atoms And Molecules Quiz 10

Which of the following statements is incorrect?

0%
One gram atom of carbon contains Avogadro's number of atoms.
0%
One mole of oxygen gas contains Avogadro's number of molecules.
0%
One mole of hydrogen gas contains Avogadro's number of atoms.
0%
One mole of electrons stands for $6.02\times{10}^{23}$ electrons.

Explanation

1 mole of any substance in a monovalent state has Avogadro's number of atoms in it which implies 1 mole of H has

$6.022 \times {10}^{23}$ of atoms in it.

Since

${H}_{2}$ gas has 2 hydrogen atoms in it, thus contains twice Avogadro's number of atoms.

Thus, 1 Mole of

${H}_{2} = 2 \times \text{Avogadro's number} = 2 \times 6.022 \times {10}^{23} = 12.044 \times {10}^{23}$ .

One gram atom means 1 mole hence one gram atom of carbon contains one mole of carbon i.e Avogadro's number of atoms.

One molle of oxygen gas contains, Avogadro's number of molecules but twice the number of oxygen atoms as it exists as

$O_2$ .

One mole of electrons means Avogadro's number of electrons or

$6.022 \times 10^{23}$ electrons.

What mass of sodium chloride will react with 34.0 g of silver nitrate to produce 17 g of sodium nitrate and 28.70 g of silver chloride if the law of conservation of mass holds good ?

0%
12.35 grams
0%
11.70 grams
0%
9.32 grams
0%
None of these

Explanation

$Nacl + AgNO_{3} \rightarrow NaNO_{3}+Agcl$

By law of conservation of max

total mass on left

By law of conservation of mass

total mass on right

$\Rightarrow x+ 3y = 17+28.7$

$\boxed{x = 11.70 gm}$

1170637_1035310_ans_bc6dc32cebff421cac048a77219c90ef.jpeg

A compound contains

$10^{-2}\%$ of phosphorus. If the atomic mass of P is

$31$ , then the molecular mass of the compound having one phosphorous atom per molecule is:

0%
$3.1\times 10^5$
0%
$3.1$
0%
$31$
0%
None of these

The Avogadro number (

$N_A$ ), is changed from

$6.022\times 10^{23}$

$mol^{-1}$ to

$6.022\times 10^{20}$

$mol^{-1}$ , this would change:

0%
the ratio of chemical species to each other in a balanced equation
0%
the ratio of elements to each other in a compound
0%
the definition of mass in units of grams
0%
the mass of one mole of carbon

Explanation

Let us consider the Avogadro number (

$N_A$ ) is changed from

$6.022 \times {10^{23}}\,mo{l^{ - 1}}\,\ to \ 6.022 \times {10^{20}}\,\,mo{l^{ - 1}}$ this would be change the mass of one mole of carbon.

Hence,

mass of one mole of carbon

$=12$ gram

or it is a mass of

$6.022 \times {10^{23}}$ atoms of carbon.

Therefore,

Now, the mass of

$6.022 \times {10^{20}}$ atoms of carbon

$=\dfrac{{12}}{{6.022 \times {{10}^{23}}}} \times 6.022 \times {10^{20}}$

$=0.012\,\,g$

So, the correct option is D.

The number of molecule in

$4.25$ g of

$NH_3$ is:

0%
$1.505\times 10^{23}$
0%
$3.01\times 10^{23}$
0%
$6.02\times 10^{23}$
0%
none of these

Explanation

Molar mass $NH_3$ = 14+3 = 17g/mol

Given mass = 4.25g

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

No of moles = 4.25/17 = 0.25 mol of $NH_3$

1 mol of $NH_3$ contains $6.022\times 10^{23}$ molecules

0.25 mol $NH_3$ contains $0.25 \times$ $6.022\times 10^{23}$ = $1.5055\times$ $10^{23}$ molecules

A sample of potato starch was ground in a ball mill to give a starch-like molecule of lower molecular weight. The product analysed was 0.086 % phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the molecular mass of the material?

0%
$3.42\times 10^4$ amu
0%
$3.72\times 10^4$ amu
0%
$3.6\times 10^4$ amu
0%
None of these

Explanation

Since the percentage of

$P$ is

$0.086$

$\therefore$

$0.086g$ of

$P$ will be present in

$100g$ of starch.

Then,

$1 g$ of

$P$ will be present in

$\dfrac{100}{0.0866}$ g of starch

The atomic mass of

$P$ is

$31 \ a.m.u$ .

$31 \ a.m.u$ of

$P$ will be contained in

$\dfrac{100\times 31}{0.0866} \ a.m.u \$ of Starch.

$\therefore$ Molecular weight of material=

$3.6 \times 10^4 \ a.m.u$

A compound contains

$4$ % oxygen, then the minimum molecular weight of that compound will be:

0%
$400g$
0%
$800g$
0%
$200g$
0%
$100g$

Explanation

$1$ mole of oxygen

$=16\ grams$

For the Molecular weight of the molecule to be least, no of oxygen moles should be

$1$ .

Let's assume that there are

$16$ gms of oxygen present in the compound, which should be

$4$ % of the total molecular mass.

$\dfrac { 4x }{ 100 } =16\ gm$

$4x =1600\ gm$

Therefore,

$x=400 \ grams$

Hence, option

$A$ is correct.

A gaseous mixture contains oxygen and nitrogen in the ratio of 1: 8 by mass. Therefore, the ratio of their respective number of molecules is:

0%
1 : 8
0%
1 : 1
0%
7 : 64
0%
1 :2

Explanation

No of Molecules =

$\dfrac{Weight}{Gram\space Molecular\space Weight} \times {N}_{A}$

Molecular wt of Hydrogen

${N}_{2}$ = 28

Molecular wt of Oxygen

${O}_{2}$ = 32

So Ratio of Number of molecules =

$\dfrac{1}{32} : \dfrac{8}{28}$ =

$7 : 64$

Mass of one atom of

$X$ is

$2.66\times 10^{-23}\ g$ , then its

$32\ g$ contains mole equal to

0%
$32\times 2.66\times 10^{-23}\ mol$
0%
$\dfrac{32}{2.66\times 10^{-23}\times 6.02\times 10^{23}}\ mol$
0%
$\dfrac{32\times 2.66\times 10^{-23}}{6.02\times 10^{23}}\ mol$
0%
$None\ of\ these$

The number of atoms of oxygen present in 10.6 g of

$Na_2CO_3$ will be:

0%
$6.02 \times 10^{22}$
0%
$12.04 \times 10^{22}$
0%
$1.806 \times 10^{23}$
0%
$31.8$

Explanation

Molar mass of

$Na_2CO_3$

$= (2 \times atomic\ mass\ of\ nitrogen) + (1 \times atomic\ mass\ of\ carbon) + (3 \times atomic\ mass\ of\ oxygen)$

$= 2 (23) + 1 (12) + 3 (16)$

$= 106\ g/mol$

So,

$1$ mole of

${ Na }_{ 2 }{ CO }_{ 3 }=108g$

Moles in

$10.6g$ of

${ Na }_{ 2 }{ CO }_{ 3 }=10.6/106=0.1$ mole

$106g$ contain atoms of oxygen

$=3\times 6.022\times { 10 }^{ 23 }$

$10.6$ contain atoms

$=0.1\times 3\times 6.022\times { 10 }^{ 23 }$

$=1.8\times { 10 }^{ 23 }\ atoms$

Which of the following is correct increasing order of molecular mass?

0%
${ H }_{ 2 }O<HCl<{ H }_{ 2 }S<{ CO }_{ 2 }$
0%
$H_2O<H_2S<HCl<CO_2$
0%
${ H }_{ 2 }O<CO_2<<HCl<{ H }_{ 2 }S$
0%
$HCl<{ H }_{ 2 }S<{ H }_{ 2 }O<{ CO }_{ 2 }$

Explanation

Atomic mass of

$H=1$

Atomic mass of

$O=16$

Atomic mass of

$Cl=35.5$

Atomic mass of

$C=12$

Atomic mass of

$S=32$

Molecular mass of the given compounds are-

$H_2O=2+16=18$

$HCl=1+35.5=36.5$

$H_2S=2+32=34$

$CO_2=12+32=44$

Hence, the correct increasing order for the molecular mass is,

$H_2O<H_2S<HCl<CO_2$

Molecules consisting of more than three atoms are called polyatomic molecules.

0%
True
0%
False

Explanation

Molecules containing more than 3 atoms are called polyatomic molecules.

Example:

$P_4$ is a polyatomic molecule.

Find number of oxygen atoms present in

$100 \ mg$ of

$CaCO_3$ .
(Atomic Mass of

$Ca=40 \ u, \ C=12 \ u, \ O=16 \$ )

0%
$6.02 \times 10^{23}$
0%
$6.02 \times 10^{20}$
0%
$1.806\times 10^{21}$
0%
$1.204 \times 10^{20}$

Explanation

Moles of

$CaCO_3=\cfrac {100}{10^3 \times 100}=1 \ milli moles$

Number of oxygen atoms in

$CaCO_3=3$

Total number of oxygen atoms in

$1 \ milimoles \ CaCO_3=10^{-3} \times 3 \times 6.02 \times 10^{23}=1.806 \times 10^{21}$

$x \ g$ of an element

$A$ contains

$\cfrac y{80}$ atoms and

$2x \ g$ of an element

$B$ contains

$\cfrac y{40}$ atoms then the ratio of atomic weights of two elements

$A$ and

$B$ is:

0%
$1:1$
0%
$1:4$
0%
$4:1$
0%
$2:1$

Explanation

Let the atomic mass of

$A$ be

$M_A$ and

$B$ be

$M_B$

Given,

$\cfrac {x}{M_A} \times N_A=\cfrac y{80}....(i)$ (

$N_A$ is avogadro number)

$\cfrac {2x}{M_B} \times N_A=\cfrac {y}{40}.....(ii)$

$(i)/(ii) \Rightarrow \cfrac {M_A}{M_B}=1:1$

Which of the following has the highest mass?

0%
(a) $50$ gms of iron
0%
(b) $5$ moles of nitrogen gas
0%
(c) $1$ gm of atom
0%
(d) $5 \times 10^{23}$ atoms of carbon

Explanation

Hint: Mass of an atom is directly proportional to number of moles.

The formula used:

Number of moles = $\dfrac {Mass} {Molar mass}$

Number of atoms = $Moles\times6.023\times10^{23}$

Step 1: The mass of iron in $50$ gram.

Here, $50$ gram already denotes the mass of iron.

Step 2: The mass of $5$ moles of nitrogen gas

According to moles formula mentioned above,

The mass of nitrogen = $5\times 28=140g$

Step 3: The mass of $1$ gm atoms.

$1$ gm atoms= $1$ mole of any atom

Step 4: The mass of $5\times10^{23}$ atoms of carbon.

Using above formula,

Mass of carbon = $\dfrac {12\times5\times10^{23}} {6.023\times10^{23}} = 9.96gm$

Among all , $5$ moles of nitrogen has highest mass of $140g$

Final answer:

Option $B$ is correct answer

If the molecular mass of a compound is

$74.5$ then the compound is:

0%
$LiCl$
0%
$HCl$
0%
$NaCl$
0%
$KCl$

Explanation

$(1 atom\times 39.0983 Potassium)+(1 atom\times 35.453 Chlorine)$ =

$(74.5513 g/mol )$ , which is the molar mass for

$KCl$ .

Hence, the answer is option

$D$ .

A sample of

${ CaCO }_{ 3 }$ has

$Ca=40$ %,

$C=12$ % and

$O=48$ %. If the law of constant proportion is true then the weight of calcium in

$5g$ of a sample of

${ CaCO }_{ 3 }$ from another source will be:

0%
$0.20g$
0%
$2.0g$
0%
$2.5g$
0%
$4.0g$

Explanation

$100g$ of sample

$\rightarrow$

$40g$ of

$Ca$

$5g$ of sample

$\rightarrow$

$\cfrac { 40 }{ 100 } \times 5=2.0\ g$ of

$Ca$

Total no. of atoms in

$44\ g$ of

$CO_{2}$ is:

0%
$6.02\times 10^{23}$
0%
$6.02\times 10^{24}$
0%
$1.806\times 10^{24}$
0%
$18.06\times 10^{22}$

Explanation

We have given mass of

$CO_2=44\ g$

Molar mass of

$CO_2=44\ g$

$44\ g\ of\ CO_2\ has\ 6.0\times 10^{23}\ molecules\ of\ CO_2$

(no. of moles

$\dfrac{44}{44}=1$ )

So, No of atoms present in

$44\ g\ of\ CO_2=6.022\times 10^{23}\ atoms$

1 molecules of

$CO_2=3\ atoms(1C+2O)$

So,

$44\ g\ of\ CO_2\ has\ 6.022\times 10^{23}\times 3=18.066\times 10^{23}\ atoms$

So, the answer is

$1.8066\times 10^{24}\ atoms$

In an experiment reproducing the measurements of Rutherford and his co-workers,

$22 \times 10^{-3}$ mg of He gas was collected in one year from a sample of radium. This sample was observed to emit

$1.06 \times 10^{11} \alpha$ - particles per second. Thus, Avogadro's number is?

0%
$5.92 \times 10^{23} mol^{-1}$
0%
$6.92 \times 10^{23} mol^{-1}$
0%
$6.08 \times 10^{23} mol^{-1}$
0%
$6.58 \times 10^{23} mol^{-1}$

Explanation

$\begin{array}{l} 22 \times 10^{-3} \text {mg per year }=22 \times 10^{-6} \mathrm{~g} \text { peryear } \\ =\frac{22 \times 10^{-6} \mathrm{~g}}{365 \times 24 \times 3600} \\ =6.976 \times 10^{-13} \mathrm{~g} \text { persecond } \\ \text { Thus, mass of } 1.06 \times 10^{11} \alpha \text { particles. } \\ \qquad \begin{aligned} =6.976 \times 10^{-13} \mathrm{~g} \\ \text { } \end{aligned} \end{array}$

$4 \mathrm{~g} \mathrm{~mole}^{-1} \quad{ }_{4}\mathrm{He}^{2}(\alpha-$ particle

$)$

$\begin{aligned}=& \frac{1.06 \times 10^{11} \times 4}{6.976 \times 10^{-13}} \\=& 6.08 \times 10^{23}\end{aligned}$

Correct answer -

$\underline{\underline{C}}$

A compound contains 3.2% of oxygen. The minimum molecular weight of the compound is:

0%
300 u
0%
440 u
0%
350 u
0%
500 u

Explanation

The minimum number of oxygen atoms the compound can contain is

$1.$

Atomic weight of oxygen=

$16 \ u$

If 3.2 % of the compound weighs

$16 \ u$

Then 100 % of the compound will weigh =

$\left(\cfrac {100}{3.2}\right)\times 16=500 \ u$

$\therefore$ Minimum molecular weight of the compound =

$500\ u$

$\overset { 14 }{ \underset { 7 }{ N } }$ if mass attributed to electron were doubled & the mass attributed to protons were halved, the atomic mass would become approximately:-

0%
Halved
0%
Doubled
0%
Reduced by 25%
0%
Remain same

Which of the following contains the largest mass of hydrogen atoms?

0%
$0.5$ moles $C_2H_2O_4$
0%
$1.1$ moles $C_3H_8O_3$
0%
$1.5$ moles $C_6H_8O_6$
0%
$4.0$ moles $C_3H_8O_3$

Explanation

1 mole of

$C_{2}H_{2}O_{4} \longrightarrow$ 2 moles of hydrogen=2 g of hydrogen

Thus 0.5 mole of

$C_{2}H_{2}O_{4}$ =1 g of hydrogen

1 mole of

$C_{3}H_{8}O_{3} \longrightarrow$ 8 moles of hydrogen=8 g of hydrogen

Thus 1.1 moles of

$C_{3}H_{8}O_{3}$ =8.8 g of hydrogen

1 mole of

$C_{6}H_{8}O_{6} \longrightarrow$ 8 moles of hydrogen=8 g of hydrogen

Thus 1.5 moles of

$C_{6}H_{8}O_{6}$ =12 g of hydrogen

1 mole of

$C_{2}H_{4}O_{2} \longrightarrow$ 4 moles of hydrogen=4 g of hydrogen

Thus 4 moles of

$C_{3}H_{8}O_{3}$ =16 g of hydrogen

Thus 4 moles of

$C_{3}H_{8}O_{3}$ contains the largest mass of hydrogen.

Hence, option

$D$ is correct.

Haemoglobin contains

$4.6\%$ of iron by mass. If the compound contains a single atom of iron then what is its molar mass in

${ g\ mol }^{ -1 }$ ?

0%
$1217.4$
0%
$1324.7$
0%
$1232.4$
0%
$1317.5$

Explanation

The percentage of Iron is

$4.6 \%$

This means that

$4.6\ g$ of Iron will be present in

$100 \ g$ of haemoglobin.

1 g of iron will be present in

$\dfrac{100}{4.6}\ g$ of haemoglobin.

Molar mass of Iron

$56 \ g \ mol^{-1}$

$56\ g \ mol^{-1}$ of iron will be present in

$\dfrac{100\times 56}{4.6} g \ mol^{-1}$ of haemoglobin.

$Molar \ mass \ of \ haemoglobin \ is =\dfrac{5600}{4.6}=1217.4\ g\ mol^{-1}$

A solution of 0.640 g of azulene in 100.0 g of benzene boils at

$80.23^0C$ . The boiling point of benzene is

$80.10^0C$ ; the

$K_b$ is

$2.53^0C/molal$ . What is the molecular weight of azulene?

0%
108
0%
99
0%
125
0%
134

Which of the following is the best example of law of conservation of mass?

[n and m are the masses of reactants and p and q are the masses of products formed.]

0%
$n-m=p-q$
0%
$n+m=p+q$
0%
$n=m$
0%
$p=q$

Explanation

Law of conservation of mass:- Mass is neither created nor destroyed it just from one compound to other.

$n+m\rightarrow$

$p+q$

Hence, the mass of reactants is equal to the mass of products formed.

Option

$B$ is correct.

Molecular mass of

${ Na }_{ 2 }{ SO }_{ 4 }$ .

$10{ H }_{ 2 }O$ is:

0%
320
0%
321
0%
322
0%
324

Explanation

$Na_2SO_4.10H_2O$

$=(2\times 23)+32+(4 \times 16)+(10\times 18)$

$=46+32+64+180$

$=322$

If Avogadro's number would have been

$1\times { 10 }^{ 23}$ , instead of

$6.02\times { 10 }^{ 23 }$ then mass of one atom of

$_8^{16}O$ would be:

0%
$1\ amu$
0%
$1\times { 10 }^{ 10 }\ amu$
0%
$16\ amu$
0%
$6\times { 10 }^{ 13 }\ amu$

Explanation

Atomic weight of

$O=16\ g/mole$

We know that

$1$ mole of

$O$ contains

$N_a=6.023\times 10^{23}$ number of atoms (Avogadro's Number of atoms)

Therefore, mass of

$6.023\times 10^{23}$ atoms of

$O=16\ g$

$\Rightarrow$ Mass of

$1$ atom of

$O=\dfrac{16\ g}{6.023\times 10^{23}}\ ....(i)$

Now,

$1\ amu = \dfrac{1}{12}th$ the mass of one

$^{12}C$ atom.

Atomic weight of carbon-12 atom

$=12\ g$

$12\ g$ of

$^{12}C$ atom contains

$6.023\times 10^{23}$ atoms.

Therefore, mass of

$1$ atom of

$^{12}C=\dfrac{12}{6.023\times 10^{23}}$

$\therefore\ \dfrac{1}{12}th$ the mass of

$1$

$^{12}C$ atom

$=1\ amu= \dfrac{1}{12}\times \dfrac{12}{6.023\times 10^{23}}=\dfrac{1}{6.023\times 10^{23}}$

Now, according to the given condition,

$N_a=1\times 10^{23}$ instead of

$6.023\times 10^{23}$ .

So,

$1\ amu$ will now be

$=\dfrac{1}{1\times 10^{23}}$

Now, putting the value of

$1\ amu$ in equation

$(i)$ , we get;

Mass of

$1$ atom of

$O=\dfrac{16}{1\times 10^{23}}\ g=16\times 10^{-23}\ g$

The value of mass of

$1$ atom of oxygen in

$amu$ :

${16}\times {10^{-23}}\times \dfrac{1}{1\times 10^{23}}\ amu$

$=16\ amu$

Hence, option (C) is correct.

Assertion: For the formation of a molecule at least two atoms are needed.
Reason: Molecules are formed by bonding between different atoms of different elements or complexes without any exception.

0%
Both Assertion and Reason are true and Reason is correct explanation
0%
Both Assertion and Reason are true and Reason is not correct explanation
0%
Assertion is correct but Reason is wrong
0%
Both Assertion and Reason are wrong

Explanation

A molecule forms when two or more atoms of the same elements join together by forming chemical bonds. So, the assertion is correct but the reason is not correct.

For example:

$Cl + Cl \rightarrow Cl_2$ (Chlorine molecule)

Hence, the correct option is C.

How many moles of helium gas occupy

$22.4L$ at

$0^\circ C$ at

$1$ atm pressure?

0%
$0.11$ mole
0%
$0.90$ mole
0%
$1.0$ mole
0%
$1.11$ mole

Explanation

$pV=nRT$

$n=\dfrac { pV }{ RT }$

$=\dfrac { 1atm\left( 22.4L \right) }{ \left( 0.082Latm/molK \right) 273K }$

$n=1$ mole

$1$ mole of the gas occupies

$22.4L$ at

${ O }^\circ C$ at

$1atm$ pressure.

Hence, the correct option is C.

Four different experiments were conducted in the following ways-

$3g$ of carbon was burnt in

$8g$ of oxygen to give

$11g$ of

$CO_2$ .

II)

$1.2g$ of carbon was burnt in air to give

$4.2g$ of

$CO_2$ .

III)

$4.5g$ of carbon was burnt in enough air to give

$11g$ of

$CO_2$ .

IV)

$4g$ of carbon was burnt in oxygen to form

$30.3g$ of

$CO_2$ .

Law of constant proportions is illustrated in which of the following experiment(s)?

0%
I and III only
0%
II and III only
0%
IV only
0%
I only

Explanation

The law of definite proportion or law of constant composition states that a given chemical compound always contains its component elements in a fixed ratio (by mass) and does not depend on its source and method of preparation.

In the option,

$A$ ,

$3g$ of carbon reacts with

$8g$ of oxygen to form

$11g$ of carbon dioxide.

$C + O_2 \xrightarrow[]{}CO_{2(g)}$

$3 + 8 = 11$

Thus, it follows the law of constant proportion. Option

$D$ is correct.

${ K }_{ 4 }\left[ Fe\left( CN \right) _{ 6 } \right]$ is supposed to be

$40$ percent dissociated when

$1M$ solution prepared. Its boiling point is equal to another

$20$ percent mass by volume of non-electrolytic solution

$A$ Considering molality=molarity . The molecular weight of

$A$ is:

0%
$77$
0%
$67$
0%
$57$
0%
$47$

A mixture of

$3\times10^{21}$ molecules of

$X$ and

$4.5\times10^{21}$ molecules of

$Y$ weigh

$1.375\ g$ . If the molecular weight of

$X$ is

$50\ g$ , then what is the molecular weight of

$Y$ ?

0%
$100$
0%
$150$
0%
$250$
0%
$300$

Explanation

Given,

Number of molecules of $X = 3\times 10^{21}$

Number of molecules of $Y = 4.5\times 10^{25}$

Molecular weight of $X = 50$

Total mass of mixture $= 1\ g$

Formula used :

$Mass\ of\ mixture=\left[\dfrac{Number\ of\ molecules\ of\ X}{Avogrado's\ Number} \times Molar\ Mass\ of\ X\right]$

$+\left[\dfrac{Number\ of\ molecules\ of\ Y}{Avogrado's\ Number}\times Molar\ Mass\ of\ Y\right]$

Now, putting all the given values in this formula, we get the molecular mass of $Y$ .

$1=\left[\dfrac{3\times 10^{21}}{6.023\times 10^{23}}\times 50\right]+\left[\dfrac{4.5\times 10^{25}}{6.023\times 10^{23}}\times Y\right]$

$\Rightarrow 1=\dfrac{150\times 10^{21}+4.5Y\times 10^{25}}{6.023\times 10^{23}}$

$\Rightarrow 6.023\times 10^{23}=10^{21}(150+4.5Y\times 10^4)$

$\Rightarrow \dfrac{6.023\times 10^{23}}{10^{21}}=150+4500Y$

$\Rightarrow 602.3=150+4500Y$

$\Rightarrow 4500Y=602.3-150$

$\Rightarrow Y = \dfrac{450.3}{4500}$

$Y= 150$

Therefore, on solving, we get $Y=150$

Therefore, the molecular weight of $Y$ is $150g$ .

Hence option (B) is correct.

A box measures

$10\ cm \times 11.2\ cm \times 10\ cm$ . Assume that this box is filled with neon gas at

$1$ atm pressure and

$273\ K$ temperature. How many electrons will be there in the box ?

0%
$6.022\times 10^{23}$
0%
$3.011\times 10^{23}$
0%
$6.022\times 10^{22}$
0%
$3.011\times 10^{22}$

The total number of oxygen atoms in

$0.2mol$ of

${Na}_{2}{B}_{4}{O}_{7}.10{H}_{2}O$ will be?

0%
$6.02\times {10}^{23}$
0%
$1.02\times {10}^{24}$
0%
$2.05\times {10}^{24}$
0%
$2.05\times {10}^{23}$

Explanation

$1\ mol$ of

$Na_{2}B_{4}O_{4} \cdot 10H_{2}O$ contains

$6.022\times 10^{23} molecules$

$0.2\ mol\ Na_{2}B_{4}O_{7}\cdot 10H_{2}O \rightarrow 0.2\times 6.022\times 10^{23} molecules$

Then, the no. of oxygen atoms

$= 0.2\times 6.022\times 10^{23}\times 17\ atoms$

$= 2.047 \times 10^{24} molecules$

Hence, Option (C) is correct.

What is the molecular mass of sodium sulphate?

0%
154
0%
119
0%
165
0%
142

If we consider that

$\frac{1}{6}$ , in place of

$\frac{1}{12}$ ; mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

0%
Decrease twice
0%
Increase two fold
0%
Remain unchanged
0%
Be a function of the molecular mass of the substance

Number of electrons in 1.8 mL of

${H}_{2}O$ are:

0%
$6.02 \times {10}^{23}$
0%
$3.01 \times {10}^{24}$
0%
$1.8 \times {10}^{23}$
0%
$6.02 \times {10}^{25}$

Explanation

There is a total of 10 electrons in 1

$H_{2}O$ Molecule (2 electrons from hydrogen and 8 electrons from oxygen.)

moles of

$H_{2}O (n) = \dfrac{mass\, of\, substance}{motor \,mass}$

18 g of

$H_2O$ = 1 mole
mass of 1.8 ml of

$H_2O= 1.8 \ ml \times 1 \ g/ml= 1.8\ g$ [

$\because Density \ of\ {H_2O} = 1 \ g/ml$ ]

1.8 g of water = 0.1 moles of water

Therefore, 1.8 ml of

$H_{2}O$ means 0.1 mole of

$H_{2}O$

Number of molecules = n

$\times$ Avogadro Number

Number of molecules =

$0.1 \times 6.022 \times 10^{23}$

$= 0.6022 \times 10^{23}$

Number of electron in total molecules of

$H_{2}O$

$= 0.6022 \times 10 ^{23} \times 10$

Number of electron in total molecules of

$H_{2}O$

$= 6.022 \times 10^{23}$

Option

$A$ is correct.

The molecular mass of

$K_2CO_3$ is:

0%
128 u
0%
112 u
0%
116 u
0%
138 u
0%
90 u

Explanation

$Molecular\ mass\ of\ K_2CO_3\ = (39 \times 2) + 12 + (16 \times 3) = 78 +12 + 48 = 138\ u$

If the atomic weight of oxygen were taken as

$90$ , then what would be the molecular weight of water?

0%
$101.25$
0%
$104.52$
0%
$112.5$
0%
$142.5$

Explanation

If the atomic weight of oxygen =

$90\ g$ .
Then the factor by which oxygens atomic mass changes is,

$\cfrac{90}{16}= 5.62$

If we go with the same assumption then the mass of hydrogen will also get change by the same factor. So, a new mass of two hydrogen atoms will be,

$5.62\times 2=11.24 g$

now, a total mass of water would be,

$90+11.24=101.25\ g$

Hence, the correct option is (A).

Avogadro's number of helium atoms have a mass of

0%
$6.023\times 10^{23}g$
0%
$4g$
0%
$8g$
0%
$4 \times\ 6.02 \times 10^{23}\ g$

Explanation

Here, Avogadro number of helium atoms means,

$6.023 \times 10^{23}$ atoms of He.

$6.023\times 10^{23}$ atoms equals to

$1$ mole of Helium.

$1$ mole of Helium has mass

$4\ g$

Hence, option

$B$ is correct.

3 molecules of

$CO_{2}$ weight

0%
$33\times10^{-23}$ g
0%
$11\times10^{-23}$ g
0%
$44\times10^{-23}$ g
0%
None of these

1 atomic mass unit is equal to:

0%
$\tfrac{1}{12}^{th}$ mass of a carbon-12 atom
0%
$1.66\times 10^{-24}\ g$
0%
$6.023\times 10^{-23}\ g$
0%
$6.023\times 10^{23}\ g$

Explanation

One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

So, we can write 1 atomic mass unit

$= \dfrac {1}{12}$ mass of carbon-12 atom

$=\dfrac {1}{12}$

$\times$

$\dfrac {12}{{6.023 \times 10^{23}}} g$

$=1.66 \times 10^{-24}\ g$

A compound was found to contain

$5.37\%$ nitrogen by mass. What is the minimum molecular weight of compound?

0%
$26.07$
0%
$2.607$
0%
$260.7$
0%
None

Who gave the definition of an element?

0%
Robert Boyle
0%
John Dalton
0%
Lavoisier
0%
Thomson

Explanation

The word element was first given by Rebert Boyle. It is defined as the simplest chemical substance that cannot be broken down during a chemical reaction.

So, the correct option is

$A$

$12$ gm of an alkaline earth metal gives

$14.8$ g of its nitride. The atomic mass of metal is:

0%
$12$
0%
$24$
0%
$20$
0%
$40$

Explanation

Let the alkaline earth metal be

$A$ .

$\therefore 3A+N_{2}\rightarrow A_{3}N_{2}$

$\Rightarrow \dfrac{Molecular\ mass\ of\ compound}{Atomic\ mass\ of\ metal}$

$=\dfrac{Wt.\ of\ compound}{Wt.\ of\ metal}$

Let the atomic mass of metal be

$x$ .

$\rightarrow$ Molecular mass of compound

$=3x+28$

$\Rightarrow \dfrac{3x+28}{3x}=\dfrac{14.8}{12}$

$\Rightarrow x=40$

A compound possesses 8% sulphur by mass. The least molecular mass is

0%
100
0%
600
0%
400
0%
200

Explanation

Hint: The minimum number of Sulphur atoms which a molecule can possess is one.

Step 1: Finding the molecular mass of the compound.

$Mass\ of\ 1\ sulphur\ atom\ =\ 32 \ grams.$

$As\ the\ compound\ has\ 8\% \ sulphur\ by\ mass,$

$\therefore 8 \ grams\ of\ sulphur\ is\ present\ in\ 100 \ grams\ of\ compound.$

$\therefore 32 \ grams\ of\ sulphur\ is\ present\ in\ \dfrac{32}{8} \times 100=400\ grams \ of\ compound.$

$Hence,\ the\ least\ molecular\ mass\ of\ the\ compound\ is\ 400 \ grams.$

Final Step: Correct answer - (C) The least molecular mass of the compound is $400 \ grams.$

A sample of pure carbon dioxide, irrespective of its source contains

$27.27 \%$ carbon and

$72.73 \%$ oxygen. The given data supports:

0%
Law of constant proportions
0%
Law of conservation of mass
0%
Law of reciprocal proporties
0%
Law of multiple proportions

Explanation

The ratio by mass in which carbon and oxygen present in

$CO_2$ :

$\dfrac{mass \ of \ carbon} {mass \ of \ oxygen}$ =

$\dfrac{12}{2×16}$

Therefore the ratio by mass in

$CO_2$ becomes

$3:8$

Now we are given the percentage of carbon and oxygen in

$CO_2$ :

Now let us find the ratio by percentage in which carbon and oxygen are present in

$CO_2$ .

$\dfrac{carbon \%} {oxygen \%} = \dfrac{27.27} {72.73}$

As both the value is divisible by we get:

$carbon: oxygen =3:8$

Carbon dioxide prepared by any method contains carbon and oxygen in the same proportion by mass. Thus, we can say that the given data proves the law of constant proportions. The Law of constant proportion is also known as the law of definite proportion.

Hence, option A is correct.

The accepted unit of atomic and molecular mass is:

0%
kilogram
0%
gram
0%
pound
0%
atomic mass unit

If Avagadro Number

${N}_{A}$ is changed from

$6.022\times{10}^{23}{mol}^{-1}$ to

$6.022\times{10}^{20}{mol}^{-1}$ this would change

0%
The ratio of elements to each other in a compound
0%
The definitiion of mass in units of grams
0%
The mass of one mole of carbon
0%
The ratio of chemical species to each other in a balanced equation

Explanation

Avogadro number:

If Avogadro number

$N_A$ is changed from

$6.022 \times 10^{23}mol^{-1}$ to

$6.022 \times 10^{23}mol^{-1}$ . This would change the mass of one mole of carbon 12g

$\rightarrow$ 12mg i.e 1 mol of carbon atom mass = 0.012gm.

Hence option

$C$ is correct.

The weight of a molecule of compound

${C_{60}}{H_{22}}$ is

0%
$1.231 \times {10^{ - 21}}g$
0%
$24 \times {10^{ - 21}}g$
0%
$5.025 \times {10^{23}}g$
0%
$16.023 \times {10^{23}}g$

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 10 - MCQExams.com

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Practice Class 9 Chemistry Quiz Questions and Answers

CBSE Questions for Class 9 Chemistry Atoms And Molecules Quiz 10 - MCQExams.com

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Practice Class 9 Chemistry Quiz Questions and Answers

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