# 256 G of Hydrogen Reacts Completely

## Atoms and Molecules – NCERT Questions

Q one.

In a reaction, five.3 g of sodium carbonate reacted with 6 thou of ethanoic acid. The products were two.2 1000 of carbon dioxide, 0.9 grand of water and viii.two g of sodium ethanoate. Prove that these observations are in agreement with the law of conservation of mass.

Sodium carbonate + ethanoic acid

sodium

ethanoate + carbon dioxide + water

SOLUTION:

Total mass of reactants = mass of sodium

carbonate + mass of ethanoic acid

= v.3 yard + 6 chiliad = xi.3 thousand

Total mass of products = mass of sodium ethanoate + mass of carbon dioxide + mass

of water

= 8.2 chiliad + two.2 g + 0.9 thou = xi.three g

Thus, the mass of reactants is equal to the mass of products, therefore the observations are in agreement with the police force of conservation of mass.

Q 2.

Hydrogen and oxygen combine in the ratio of i : 8 by mass to grade water. What mass of oxygen gas would exist required to react completely with 3 chiliad of hydrogen gas?

SOLUTION:

ane g of hydrogen reacts with = viii m of oxygen

∴ 3 yard of hydrogen reacts with = 8 × 3 = 24 1000 of oxygen

Thus, 24 thou of oxygen gas would be required to react completely with three thousand of hydrogen gas.

Q iii.

Which postulate of Dalton’southward atomic theory is the result of the police force of conservation of mass?

SOLUTION:

The postulate that “atoms can neither be created nor destroyed in a chemical reaction” is the result of the law of conservation of mass.

Q four.

Which postulate of Dalton’s atomic theory tin can explain the police of definite proportions?

SOLUTION:

The postulate that “A chemical compound always consists of the same elements combined together in the same proportion by mass” is the law of definite proportions.

Q 5.

Ascertain the atomic mass unit.

SOLUTION:

Atomic mass unit is defined every bit the mass unit of measurement equal to exactly one-twelfth (i/12th) of the mass of 1 atom of carbon-12. Information technology is denoted by u (unified mass).

i.eastward. 1 u = 1.66 × 10–24
g

Q vi.

Why is it non possible to see an atom with naked eyes?

SOLUTION:

Information technology is not possible to come across an cantlet with naked eye because of its extremely small size (diminutive radius is of the order of 10–10
thousand).

Q 7.

Write down the formulae of

(I) sodium oxide,

(ii) aluminium chloride,

(3) sodium sulphide,

(4) magnesium hydroxide

SOLUTION:

(I) Sodium oxide

(two) Aluminium chloride

(three) Sodium sulphide

(iv) Magnesium hydroxide

Q 8.

Write downwards the names of compounds represented by the post-obit formulae :

(I) Al2(And theniv)3

(two) CaCl2

(3) K2SO4

(iv) KNO3

(five) CaCO3

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SOLUTION:

(I) Aluminium sulphate

(ii) Calcium chloride

(iii) Potassium sulphate

(iv) Potassium nitrate

(five) Calcium carbonate

Q 9.

What is meant by the term chemic formulae?

SOLUTION:

The chemical formula of a chemical compound is a symbolic representation of its composition.
due east.thou.,
formula of calcium oxide.

Thus, chemical formula of calcium oxide is CaO.

Q x.

How many atoms are present in a

(I) H2Due south molecule and

(ii) POiv
3–
ion?

SOLUTION:

(I) iii atoms considering HiiDue south molecule has two atoms of hydrogen and 1 atom of sulphur.

(ii) 5 atoms because

ion has one atom of phosphorus and 4 atoms of oxygen.

Q xi.

Calculate the molecular masses of Htwo, Oii, Cl2, COtwo, CH4, C2Hsix, CiiHfour, NHthree, CH3OH

SOLUTION:

Molecular mass of H2
= two × atomic mass of H

= two × 1 u = ii u

Molecular mass of O2
= ii × atomic mass of O

= 2 × sixteen u = 32 u

Molecular mass of Cl2
= 2 × atomic mass of Cl

= 2 × 35.5 u = 71 u

Molecular mass of CO2
= diminutive mass of C

+ 2 × atomic mass of O

= 12 + (ii × 16) = (12 + 32) u = 44 u

Molecular mass of CHfour
= atomic mass of C

+ iv × atomic mass of H

= 12 + (4 × i) u = (12 + 4) u = 16 u

Molecular mass of C2H6
= ii × atomic mass

of C + 6 × atomic mass of H

= (two × 12 + 6 × ane) u = (24 + half dozen) u = 30 u

Molecular mass of CtwoHiv
= 2 × atomic mass

of C + 4 × atomic mass of H

= (2 × 12 + iv × 1) u = (24 + 4) u = 28 u

Molecular mass of NH3
= atomic mass of N

+ 3 × atomic mass of H

= (14 + three × 1) u = (14 + 3) u = 17 u

Molecular mass of CH3OH = atomic mass of C

+ 3 × atomic mass of H + diminutive mass

of O + atomic mass of H

= (12 + 3 × 1 + 16 + 1) u = (12 + 3 + 17) u = 32 u

Q 12.

Calculate the formula unit masses of ZnO, NatwoO, One thousandiiCO3. Given atomic masses of Zn = 65 u, Na = 23 u, M = 39 u, C = 12 u and O = xvi u

SOLUTION:

Formula unit of measurement mass of ZnO = atomic mass of Zn

+ diminutive mass of O

= (65 + 16) u = 81 u

Formula unit mass of NaiiO = two × atomic mass

of Na + atomic mass of O

= (2 × 23 + xvi) u = 62 u

Formula unit mass of KiiCO3

= 2 × atomic mass of Thousand + atomic mass of C

+ 3 × diminutive mass of O

= (2 × 39 + 12 + 3 × 16) u = (78 + 12 + 48) u

= 138 u

Q 13.

If i mole of carbon atoms weighs

12 gram, what is the mass (in gram) of

ane atom of carbon?

SOLUTION:

Molecular mass of carbon = 12 g

∵ six.022 × 1023
atoms of carbon have mass

= 12 k

∴ 1 atom of carbon has mass

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Q 14.

Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given diminutive mass of Na = 23 u, Atomic number 26 = 56 u)?

SOLUTION:

100 g of sodium :

Number of sodium atoms

Thus, 100 thousand of sodium has more atoms.

Q xv.

A 0.24 one thousand sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 k of oxygen. Calculate the percentage composition of the compound by weight

SOLUTION:

We know that, % of whatsoever chemical element in a compound

Q 16.

When 3.0 yard of carbon is burnt in 8.00 one thousand oxygen, 11.00 chiliad of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 chiliad of carbon is burnt in 50.00 yard of oxygen? Which law of chemical combination will govern your answer?

SOLUTION:

Total mass of reactants = mass of C + mass

of O2
= 3 + 8 = 11 1000

Total mass of reactants = total mass of

products

Hence, the law of conservation of mass is proved.

Further, it also shows carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is three : 8. Thus, it also proves the law of constant proportions. 3 1000 of carbon must likewise combine with 8 g of oxygen merely. This ways that (fifty – viii) = 42 g of oxygen volition remain unreacted.

Q 17.

What are polyatomic ions? Give examples

SOLUTION:

A polyatomic ion is a group of atoms carrying positive or negative charge.

Q xviii.

Write the chemic formulae of the following :

(A)
Magnesium chloride

(B)

Calcium oxide

(C)

Copper nitrate

(D)
Aluminium chloride

(East)
Calcium carbonate

SOLUTION:

(A) Magnesium chloride

(B)
Calcium oxide

(C) Copper nitrate

(D) Aluminium chloride

(E) Calcium carbonate

Q nineteen.

Requite the names of the elements nowadays in the following compounds.

(A)
Quick lime

(B)
Hydrogen bromide

(C)
Blistering powder

(D)
Potassium sulphate

SOLUTION:

(A) Quick lime is CaO. Elements present are calcium and oxygen.

(B) Hydrogen bromide is HBr. Elements present are hydrogen and bromine.

(C) Baking pulverisation is NaHCO3. Elements present are sodium, hydrogen, carbon and oxygen.

(D) Potassium sulphate is K2SOfour. Elements nowadays are potassium, sulphur and oxygen.

Q 20.

Summate the tooth mass of the post-obit substances.

(A)
Ethyne, C2Hii

(B)
Sulphur molecule, Southward8

(C)
Phosphorus molecule, Pfour
(atomic mass of phosphorus = 31)

(D)
Muriatic acid, HCl

(East)
Nitric acrid, HNO3
.

SOLUTION:

(A) Molar mass of CiiH2
= ii × atomic mass of C + 2 × atomic mass of H = (two × 12 + 2 × 1) u = (24 + ii) u = 26 u

(B) Molar mass of S8
= 8 × atomic mass of Southward = (8 × 32) u = 256 u

(C) Tooth mass of Piv
= 4 × atomic mass of P = (4 × 31) u = 124 u

(D) Tooth mass of HCl = diminutive mass of H + atomic mass of Cl = (1 + 35.v) u = 36.5 u

(E) Tooth mass of HNO3
= atomic mass of H+ atomic mass of N + 3 × atomic mass of O = (ane + 14 + iii × 16) u = (1 + 14 + 48) u = 63 u

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Q 21.

What is the mass of

(A)
1 mole of nitrogen atoms?

(B)
4 moles of aluminium atoms (atomic mass of aluminium = 27)?

(C)
ten moles of sodium sulphite (NaiiAnd then3)?

SOLUTION:

(A) Mass of nitrogen atom

= number of moles × atomic mass

= i × fourteen = fourteen 1000

(B) Mass of aluminium atom

= iv × diminutive mass = 4 × 27 = 108 g

(C) Mass of Na2SOiii
= 2 × 23 + 32 + 16 × three

= 46 + 32 + 48 = 126 g

1 mole of NaiiSO3
= 126 g

∴ 10 moles of Na2Soiii
= (x × 126) 1000

= 1260 g

Q 22.

Convert into mole.

(A)
12 1000 of oxygen gas

(B)
20 one thousand of water

(C)
22 g of carbon dioxide

SOLUTION:

(A) Molar mass of O2
= 2 × 16 = 32 thou

∴ Number of moles in 12 grand of O2

(B) Tooth mass of HiiO = two × 1 + sixteen = eighteen m

∴ Number of moles in 20 g of HiiO

(C) Molar mass of CO2
= 12 + sixteen × 2

= 12 + 32 = 44 g

∴ Number of moles in 22 g of COii

Q 23.

What is the mass of

(A)
0.2 mole of oxygen atoms?

(B)
0.5 mole of h2o molecules?

SOLUTION:

(A) Mass of 1 mole of oxygen atom = 16 u

Mass of 0.2 mole of oxygen atoms

= (16 × 0.2) u = 3.ii u

(B) Mass of i mole of H2O molecule

= (one × 2 + 16) = xviii u

Mass of 0.5 mole of H2O molecule

= (18 × 0.five) u = 9 u

Q 24.

Summate the number of molecules of sulphur (S8) present in 16 thousand of solid sulphur?

SOLUTION:

Mass of 1 mole of Seight
= 8 × 32 = 256 thousand

∴ 256 g of Southviii
contains = 6.022 × x23

molecules

∴ 16 1000 of Southward8
contains

Q 25.

Calculate the number of aluminium ions present in 0.051 chiliad of aluminium oxide.

(Hint : The mass of an ion is the same equally that of an atom of the same element. Atomic mass of Al = 27 u)

SOLUTION:

Molar mass of Al2O3
= ii × 27 + 3 × xvi

= 54 + 48 = 102 g

∴ 102 g of AltwoO3
contains

= ii × 6.022 × 1023
Aliii+
ions

∴ 0.051 g of AltwoOthree
contains

= 6.022 × 1020
Althree+
ions.

## 256 G of Hydrogen Reacts Completely

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