Q: Consider the following pairs,I. CH4,C2H6II. CO,CO2III. NO,NO2IV. H2O,H2O2In which cases, the law of multiple proportions is followed?

I: C:H mass ratio in C2H6C:H mass ratio in CH4=4gc/gH3gc/gH=3:4II: C:O mass ratio in CO2C:O mass ratio in CO=6 gC/32 gO6 gC/16 gO=2:1III: N:O mass ratio in NO2N:O mass ratio in NO=14gN/32 gO14 g N/10 gO=2:1IV: H:O mass ratio in H2O2H:O mass ratio in H2O=2gH/32 gO2gH/10 gO=2:1Thus, in all cases law of multiple proportion is followed.

Q: In which case purity of the substance is 100% ?

(a) 1 mol3CaCO3⟶ΔCaO+22.4 LCO2actual =11.2 LThus, 50%(b) 1 mol3MgCO3⟶Δ40 gMgO+CO2Thus, 100%(c) 1mol2mol2NaHCO3△Na2CO3+18g9gH2O+CO2actual =4 gThus, 44.4%(d) 1 mol3Ca(HCO2⟶ΔCaO+H2O+2 mol2CO2 actual =1 mol Thus, 50%

Q: Consider the following laws of chemical combination with examplesI. Law of multiple proportion : N2O,NO,NO2II. Law of reciprocal proportion : H2O,SO2,H2 SWhich is correct with examples?

Ratio of H and S in H2 S:2 gH/32 g SRatio of H and O in H2O:2 gH/16 gORatio of S and O in SO2:32 g S/32 gOThus, law of reciprocal proportion is followed.Thus, I and II both.

Q: H2 S contains 94.11% sulphur; SO2 contains 50% oxygen and H2O contains 11.11% hydrogen. Thus,

H2 S5.89 gH combines with =94.11gShence, 1 gH combines with =16 g SSO250 gO combine with =50 g Shence, 1 gO combines with =1 g SH2O11.11 gH combines with =88.89 gO1 gH combine with =8 gOThus, law of reciprocal proportion is followed.

Q: M2+ ion is isoelectronic of SO2 and has (Z+2) neutrons ( Z is atomic number of M ). Thus, the ionic mass of M2+ is

M2+ has electrons =32 (isoelectronic of SO2protons =34=Zneutrons =36Thus,ionic mass =70

Question 1

In the decomposition of 10 g of MgCO3​,0.1 moleCO2​ and 4.0 gMgO are obtained. Hence, percentage purity of MgCO3​ is

Accepted Answer

Thus, % of pure MgCO3​=84%

Question 2

Consider the following pairs,I. CH4​,C2​H6​II. CO,CO2​III. NO,NO2​IV. H2​O,H2​O2​In which cases, the law of multiple proportions is followed?

Accepted Answer

I: C:H mass ratio in C2​H6​C:H mass ratio in CH4​​=4gc/gH3gc/gH​=3:4II: C:O mass ratio in CO2​C:O mass ratio in CO​=6 gC/32 gO6 gC/16 gO​=2:1III: N:O mass ratio in NO2​N:O mass ratio in NO​=14gN/32 gO14 g N/10 gO​=2:1IV: H:O mass ratio in H2​O2​H:O mass ratio in H2​O​=2gH/32 gO2gH/10 gO​=2:1Thus, in all cases law of multiple proportion is followed.

Question 3

Two substances I and II of carbon and oxygen have respectively 72.73% and 47.06% oxygen. Hence, they follow

Accepted Answer

The desired ratio in the given compounds is,

Question 4

In which case purity of the substance is 100% ?

Accepted Answer

(a) 1 mol3​CaCO3​⟶Δ​​CaO+22.4 LCO2​​actual =11.2 LThus, 50%(b) 1 mol3​MgCO3​⟶Δ​​40 gMgO​+CO2​Thus, 100%(c) 1mol2mol​2NaHCO3​​△​Na2​CO3​+18g9g​H2​O​+CO2​actual =4 gThus, 44.4%(d) 1 mol3​Ca(HCO2​​⟶Δ​CaO+H2​O+2 mol2CO2​​ actual =1 mol Thus, 50%

Question 5

Consider the following laws of chemical combination with examplesI. Law of multiple proportion : N2​O,NO,NO2​II. Law of reciprocal proportion : H2​O,SO2​,H2​ SWhich is correct with examples?

Accepted Answer

Ratio of H and S in H2​ S:2 gH/32 g SRatio of H and O in H2​O:2 gH/16 gORatio of S and O in SO2​:32 g S/32 gOThus, law of reciprocal proportion is followed.Thus, I and II both.

Question 6

H2​ S contains 94.11% sulphur; SO2​ contains 50% oxygen and H2​O contains 11.11% hydrogen. Thus,

Accepted Answer

H2​ S5.89 gH combines with =94.11gShence, 1 gH combines with =16 g SSO2​50 gO combine with =50 g Shence, 1 gO combines with =1 g SH2​O11.11 gH combines with =88.89 gO1 gH combine with =8 gOThus, law of reciprocal proportion is followed.

Question 7

Sodium combines with 1735​Cl and 1737​Cl to give two samples of sodium chloride. Their formation follows the law of

Accepted Answer

There are two types of NaCl formed. They differ in molar masses due to different isotopes of Cl.

Question 8

According to Dalton's atomic theory, the smallest particle in which matter can exist, is called(a) an atom(b) an ion(d) a molecule(c) an electron

Accepted Answer

(a) By Dalton's theory, atom is the smallest particle.

Question 9

The nucleus of an atom consists of

Accepted Answer

Nucleus consists of proton and neutron and molar mass = neutron + proton

Question 10

Ionic mass of X3− is 17 . If it has 10 electrons, then number of neutrons are

Accepted Answer

X3− has 10 electrons.Thus, protons ( = atomic number) in X3−=7neutron + proton = ionic mass =17Thus,  neutrons =17−7=10

Question 11

M2+ ion is isoelectronic of SO2​ and has (Z+2) neutrons ( Z is atomic number of M ). Thus, the ionic mass of M2+ is

Accepted Answer

M2+ has electrons =32 (isoelectronic of SO2​protons =34=Zneutrons =36Thus,ionic mass =70

Question 12

X+,Y2+ and Z−are isoelectronic of CO2​. Increasing order of protons in X+,Y2+ and Z− is

Accepted Answer

Number of electrons in CO2​=6+16=22Thus, increasing order of proton is Z−

Question 13

Which has maximum number of H-atoms per gram of the substance?

Accepted Answer

The maximum number of hydrogen is present in methane.

Question 14

If each O-atom has two equivalents, volume of one equivalent of O2​ gas at STP is

Accepted Answer

1 mole O2​=2 O-atoms =4 equivalents oxygenVolume of 1 mole =22.4 L at STPVolume of 1 equivalent =5.6 L at STP

Question 15

Mass of one atom of an element is 6.64×10−23 g. This is equal to

Accepted Answer

Mass of one atom =6.64×10−23 gThus, atomic mass =6.64×10−23×6.02×1023=40

Question 16

If Avogadro's number would have been 1×10−10, instead of 6.02×1023 then mass of one atom of H would be

Accepted Answer

The mass of hydrogen depends only on the proton's mass and the proton's mass is 1 amu or 1.662×10−24 g.

Question 17

Number of mole of 612​C in 1 amu is

Accepted Answer

1 mol 612​C=12 g1amu=N0​1​Number of moles in amu =12N0​1​

Question 18

Mass of one 714​ N-atom is

Accepted Answer

N0​ atoms =14 g1 atom =N0​14​=14amu

Question 19

If two compounds have same empirical formula but different molecular formula, they must have

Accepted Answer

Same empirical formula, it means ratio of atoms is identical. Hence, they differ in molecular weight.

Question 20

If the equivalent weight of an element is 32 , then the percentage of oxygen in its oxide is

Accepted Answer

Equivalent weight of element =32 gand that of oxygen =8 gThus, one equivalent of oxide =40 gPercentage of oxygen in oxide =408​×100=20%

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