33 + Interview Questions in MOLE CONCEPT IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Calculate the following :Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]
Answer» 1 mole of CO₂ has mass = 12 + (16 x 2 ) = 44g 1 mole of CO has mass = 12 + 16 = 28 g ∴ Mass of 1 mole of CO₂has more mass.

1.

Calculate the following :Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]

Answer»

1 mole of CO₂ has mass = 12 + (16 x 2 ) = 44g

1 mole of CO has mass = 12 + 16 = 28 g

∴ Mass of 1 mole of CO₂has more mass.

Discussion

2.	A vessel contains X number of molecules of hydrogen gas at a certain temperature & pressure. Under the same conditions of temperature & pressure, how many molecules of nitrogen gas would be present in the same vessel.
Answer» According to Avogadro’s law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules. Hence, number of molecules of N₂ = Number of molecules of H₂ = X

2.

A vessel contains X number of molecules of hydrogen gas at a certain temperature & pressure. Under the same conditions of temperature & pressure, how many molecules of nitrogen gas would be present in the same vessel.

Answer»

According to Avogadro’s law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N₂ = Number of molecules of H₂ = X

Discussion

3.	Select the correct answer from A, B, C and D : The ratio between the number of molecules in 2g of hydrogen and 32g of oxygen is: k (A) 1 : 2 (B) 1 : 0.01 (C) 1:1 (D) 0.01 : 1 [H = 1, O = 16]
Answer» The molecules in 2g of hydrogen and 32g of oxygen is : 1 : 1.

Discussion

4.	Explain the following :Is it possible to change the temperature and pressure of a fixed mass of gas without changing its volume ?
Answer» Yes, an increase in temperature produce an increase in volume which can be reduced (changed) to original volume by the increase in pressure.

Discussion

5.	Explain the following :One mole of hydrogen contains 2 x 6.023 x 1023 atoms of hydrogen where as one mole of helium contains 6.023 x 1023 atoms of helium
Answer» Hydrogen is a diatomic gas. So one molecule of hydrogen = 2 atoms ∴ 1 mole or 6.023 x 10²³ molecules of H₂ = 2 x 6.023 x 10²³ atoms On the other hand, helium is monoatomic gas, ∴ One molecule of helium 1 atom of He or 6.023 x 10²³ molecules of He = 6.023 x 10²³ atoms of He.

5.

Explain the following :One mole of hydrogen contains 2 x 6.023 x 1023 atoms of hydrogen where as one mole of helium contains 6.023 x 1023 atoms of helium

Answer»

Hydrogen is a diatomic gas. So one molecule of hydrogen = 2 atoms

∴ 1 mole or 6.023 x 10²³ molecules of H₂ = 2 x 6.023 x 10²³ atoms

On the other hand, helium is monoatomic gas,

∴ One molecule of helium 1 atom of He

or 6.023 x 10²³ molecules of He = 6.023 x 10²³ atoms of He.

Discussion

6.	A gas cylinder can hold 1 kg. of H2 at room temp. & press: State the law that helped you to arrive at the above result.
Answer» Avogadro’s law.

Discussion

7.	what volume (in ml)of 0.2M H2SO4 solution should be mixed with the 40ml of 0.1M NaOH SOLUTIONsuchthat the resulting solution has concentration of H2SO4 as (6/55)M
Answer» Solution: We know that resultant molarity is given by Resultant Molarity = (M₁V₁+M₂V₂)/(V₁+V₂) we have M₁ = 0.1, V₁ = 40ml = 0.04 litre M₂ = 0.2, V₂ = ? Now putting the values in above formula we get, (0.1x0.04 + 0.2xV₂)/0.04+V₂ = 6/55 => 0.004+0.2V₂ = (0.24+6V₂)/55 => 0.22+11V₂ = 0.24+6V₂ => 5V₂ = 0.02 => V₂ = 0.004 litre Hence, V₂ = 0.004 litre or 4ml

7.

what volume (in ml)of 0.2M H2SO4 solution should be mixed with the 40ml of 0.1M NaOH SOLUTIONsuchthat the resulting solution has concentration of H2SO4 as (6/55)M

Answer»

Solution: We know that resultant molarity is given by

Resultant Molarity = (M₁V₁+M₂V₂)/(V₁+V₂)

we have M₁ = 0.1, V₁ = 40ml = 0.04 litre

M₂ = 0.2, V₂ = ?

Now putting the values in above formula we get,

(0.1x0.04 + 0.2xV₂)/0.04+V₂ = 6/55

=> 0.004+0.2V₂ = (0.24+6V₂)/55

=> 0.22+11V₂ = 0.24+6V₂

=> 5V₂ = 0.02

=> V₂ = 0.004 litre

Hence, V₂ = 0.004 litre or 4ml

Discussion

8.

A compound is formed by 24 grams of X and 64 grams of oxygen, if X = 12 and O = 16. Find the simplest formula of the compound. ”

Answer»

Relative number of atoms	Simplest ratio
X = 24/12 = 2	X = 2/2 = 1
O = 64/16 = 4	O = 4/2 = 2

Therefore, simple ratio between X and O is X : O = 1 : 2

Thus, empirical formula of the compound is XO₂ .

Discussion

9.	What is the important information given by a balanced chemical equation of decomposition of hydrogen peroxide ?
Answer» The decomposition of hydrogen peroxide is given by the equation: 2H₂O₂ —> 2H₂O + O₂ The equation gives the following important information: (i) Hydrogen peroxide decomposes to form water and oxygen. (ii) Two molecules of hydrogen peroxide when decomposed, form two molecules of water and one molecule of oxygen. (iii) 68 gm of hydrogen peroxide when decomposed, yield 36 gm of water and 32 gm of oxygen

9.

What is the important information given by a balanced chemical equation of decomposition of hydrogen peroxide ?

Answer»

The decomposition of hydrogen peroxide is given by the equation:

2H₂O₂ —> 2H₂O + O₂

The equation gives the following important information:

(i) Hydrogen peroxide decomposes to form water and oxygen.

(ii) Two molecules of hydrogen peroxide when decomposed, form two molecules of water and one molecule of oxygen.

(iii) 68 gm of hydrogen peroxide when decomposed, yield 36 gm of water and 32 gm of oxygen

Discussion

10.	What are the limitations of a chemical equation?
Answer» Following are the limitations of a chemical equation: (i) A chemical equation does not tell us the physical state of the reactants and the products in the reaction. (ii) It does not tell us the actual concentration or dilution of the reactants used in the reaction. (iii) It does not tell whether the reaction starts at its own or some heat is required to start the reaction. (iv) It does not tell whether the reaction is violent in nature or not. (v) The time taken by the reaction to complete itself is also not known.

10.

What are the limitations of a chemical equation?

Answer»

Following are the limitations of a chemical equation:

(i) A chemical equation does not tell us the physical state of the reactants and the products in the reaction.

(ii) It does not tell us the actual concentration or dilution of the reactants used in the reaction.

(iii) It does not tell whether the reaction starts at its own or some heat is required to start the reaction.

(iv) It does not tell whether the reaction is violent in nature or not.

(v) The time taken by the reaction to complete itself is also not known.

Discussion

11.	“When stating the volume of a gas, the pressure and temperatrue should be also given.” Why ?
Answer» It is because, the volume of a gas changes, if the pressure or temperature or both change. Thus, while stating the stating the volume of a gas pressure and volume has to be specified along with volume.

Discussion

12.	The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.Which sample contains the maximum number of molecules. If the temp, and pressure of gas A are kept constant, then what will happen to the volume of A when the no. of molecules is doubled.
Answer» The sample D and volume of A will get Doubled.

Discussion

13.	The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.If this ratio of gas vols. refers to reactants and products of reaction – gas law observed is ….
Answer» Gay Lussac’s law of combining volumes.

Discussion

14.	Determines the relation between molecular weight and vapour density.
Answer» Molecular weight: It is the ratio of the weight of 1 molecule of a substance to the weight of one atom of hydrogen. Mol.wt = mass of 1 molecule of a substance/mass of 1 mol. of hydrogen Vapour Density : It is the ratio of the mass of a certain volume of gas or vapour of the mass of the same volume of hydrogen. Mol. wt. = 2 x vapour density

14.

Determines the relation between molecular weight and vapour density.

Answer»

Molecular weight: It is the ratio of the weight of 1 molecule of a substance to the weight of one atom of hydrogen.

Mol.wt = mass of 1 molecule of a substance/mass of 1 mol. of hydrogen

Vapour Density : It is the ratio of the mass of a certain volume of gas or vapour of the mass of the same volume of hydrogen.

Mol. wt. = 2 x vapour density

Discussion

15.	What is ‘mole scale’ of a compound?
Answer» The molecular weight of a compound expressed in gram is known as “a mole” of a compound. The multiples of the fractions of a mole give different mole values of a compound. The molecular weight of carbon dioxide is 44, hence 44 gm is one mole of carbon dioxide. 88 gm, and 176 gm of carbon dioxide represent 2 moles and 4 moles respectively. In general, Number of moles of a compound = mass of the compound in gm/Molecular weight of the compound

15.

What is ‘mole scale’ of a compound?

Answer»

The molecular weight of a compound expressed in gram is known as “a mole” of a compound. The multiples of the fractions of a mole give different mole values of a compound. The molecular weight of carbon dioxide is 44, hence 44 gm is one mole of carbon dioxide. 88 gm, and 176 gm of carbon dioxide represent 2 moles and 4 moles respectively. In general,

Number of moles of a compound = mass of the compound in gm/Molecular weight of the compound

Discussion

16.	State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if its V.D. is 8.
Answer» V.D. of hydrocarbon = 8 ∴ Molecular weigh = 2 x V.D. = 2 x 8 = 16 g 16 g of hydrocarbon occupy at S.T.P. = 22.4.lit ∴ 40 g of hydrocarbon occupy at S.T.P = 22.4 x 40/16 = 56 lits.

16.

State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if its V.D. is 8.

Answer»

V.D. of hydrocarbon = 8

∴ Molecular weigh = 2 x V.D. = 2 x 8 = 16 g

16 g of hydrocarbon occupy at S.T.P. = 22.4.lit

∴ 40 g of hydrocarbon occupy at S.T.P

= 22.4 x 40/16 = 56 lits.

Discussion

17.	State Avogadro’s Law. A cylinder contains 68g of ammonia gas at s.t.p. 1. What is the volume occupied by this gas? 2. How many moles and how many molecules ammonia are present in the cylinder? [N = 14, H = 1]
Answer» Avogadro's law states that 'equal vol.of all gases under similar conditions of temperature and pressure contain the same no.of molecules.' (i) Molecular gram atom wt. of NH₃ = 17 gm 17 gm of NH₃ has vol. at s.t.p. = 22.4 lt. ∴ 68 gm of NH₃ has vol. at s.t.p = 22.4 x 68 /17 = 89.6 it. (ii) No. of moles in 68 g of NH₃ = 68/17 = 4 moles. ⇒ 1 mole of ammonia = 6.23 x 10²³ molecules ⇒ 4 moles of ammonia = 4 x 6.023 x10²³ =24.092 x 10²³molecules

17.

State Avogadro’s Law. A cylinder contains 68g of ammonia gas at s.t.p. 1. What is the volume occupied by this gas? 2. How many moles and how many molecules ammonia are present in the cylinder? [N = 14, H = 1]

Answer»

Avogadro's law states that 'equal vol.of all gases under similar conditions of temperature and pressure contain the same no.of molecules.'

(i) Molecular gram atom wt. of NH₃ = 17 gm

17 gm of NH₃ has vol. at s.t.p. = 22.4 lt.

∴ 68 gm of NH₃ has vol. at s.t.p = 22.4 x 68 /17 = 89.6 it.

(ii) No. of moles in 68 g of NH₃

= 68/17 = 4 moles.

⇒ 1 mole of ammonia = 6.23 x 10²³ molecules

⇒ 4 moles of ammonia = 4 x 6.023 x10²³

=24.092 x 10²³molecules

Discussion

18.	The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OWhat volume, at s.t.p., is occupied by the number of moles determined in (1) (i) ?
Answer» 1 moles occupy a volume of 22.4 litres. 8 mole will occupy a volume of = 22.4 x 8 = 179.2 litres.

18.

The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OWhat volume, at s.t.p., is occupied by the number of moles determined in (1) (i) ?

Answer»

1 moles occupy a volume of 22.4 litres.

8 mole will occupy a volume of = 22.4 x 8 = 179.2 litres.

Discussion

19.	What is the volume (measured in dm3 or litres occupied by one mole of gas at S.T.P. ?
Answer» One mole of gas occupies 22.4 litre at S.T.P.

Discussion

20.	The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.
Answer» 5.6 dm³ of gas weighs = 12 g 22.4 dm³of gas weight =(12.0/5.6 x 22.4) gm = 48 gm

20.

The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

Answer»

5.6 dm³ of gas weighs = 12 g

22.4 dm³of gas weight

=(12.0/5.6 x 22.4) gm

= 48 gm

Discussion

21.	The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.If the volume of ‘A’ is 5.6 dm3 at s.t.p., calculate the no. of molecules in the actual vol. of ‘D’ at s.t.p. (Avog no. is 6 x 1023 ). Using your answer, state the mass of ‘D’ if the gas is “N2O” (N = 14, O = 16). [6 x 1023, 44g.]
Answer» Vol. of D will be 4 x 5.6 = 22.4 lit. and 22.4 lit. of D contain molecules = 6 x10²³ (AV. number) Mass of N₂ O (i.e. D) = (14 × 2 + 16 x 1) = 44g.

Discussion

22.	The vapour density of carbon dioxide [C = 12, O = 16] is (A) 32 (B) 16 (C) 44 (D) 22
Answer» The vapour density of carbon dioxide 22.

Discussion

23.	A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2g, hence the relative molecular mass of the gas is : (A) 5 (B) 10 (C) 15 (D) 20
Answer» The relative molecular mass of the gas is 10.

Discussion

24.	Why relative atomic mass is compared with 1 /12 of carbon?
Answer» Naturally occurring hydrogen has three isotopes (₁H¹ , ₁H² and ₁H³ ) and its relative atomic mass works out 1.008 rather than 1. Atomic mass of carbon is 12. Thus, 1/12 mass of carbon work out as 1. It is on account of the above reason 1/12 mass of carbon is used for comparing relative atomic masses of other elements.

Discussion

25.	Why is the term relative atomic mass used for atomic mass of an element ?
Answer» Since the actual mass of an atom of element is extremely small, for comparing the masses of atoms of different elements, the mass of an atom of some light element is fixed as standard and the masses of other atoms are expressed relative to the standard mass, so the term relative atomic mass is used.

Discussion

26.	Define the term atomic weight :
Answer» Atomic weight : is the number of times one atom of an element is heavier than 1/2 the mass of an atom of carbon (C¹² )

Discussion

27.	Mention the term defined by the following : The mass of a given volume of gas compared to the mass of an equal volume of hydrogen.
Answer» Vapour Density.

Discussion

28.	From A, B, C, D, which weighs the least — A : 2 g. atoms of Nitrogen B : 1 mole of Silver C : 22.4 litres of oxygen gas at 1 atmospheric pressure and 273K D : 6.02 x 1023 atoms of carbon. [Ag = 108, N=14, 0=16, C=12]
Answer» From A, B, C, D,weighs the least 6.02 x 10²³ atoms of carbon.

Discussion

29.	The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OIf the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?
Answer» From equation, we know that 2moles of octane produces 16 moles of CO₂. Mass of CO₂ produced = 16 x 44 = 704

29.

The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OIf the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?

Answer»

From equation, we know that 2moles of octane produces 16 moles of CO₂.

Mass of CO₂ produced = 16 x 44 = 704

Discussion

30.	A compound having empirical formula X2 Y is made of two elements X and Y. Find its molecular formula. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density 25.
Answer» Empirical formula =X₂Y, Empirical mass = 2 x 10+ 5 = 25 Vapour density of X₂ Y = 25 Molecular weight of X₂ Y = 2 x vapour density = 2 x 25 = 50 Now molecular weight =n x empirical weight = 50 = n x 25 n = 50/25 = 2 Molecular formula = n x Empirical formula = 2x X₂ Y = X₄ Y₂

30.

A compound having empirical formula X2 Y is made of two elements X and Y. Find its molecular formula. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density 25.

Answer»

Empirical formula =X₂Y, Empirical mass = 2 x 10+ 5 = 25

Vapour density of X₂ Y = 25

Molecular weight of X₂ Y = 2 x vapour density = 2 x 25 = 50

Now molecular weight =n x empirical weight = 50 = n x 25

n = 50/25 = 2

Molecular formula = n x Empirical formula = 2x X₂ Y = X₄ Y₂

Discussion

31.	How does Avogadro’s law explain Gay-Lussac’s law of gaseous volumes ?
Answer» Avagadro’s law states that equal volumes of all gases contain equal number of molecules under similar conditions of temperature and presure. Since, when gases react chemically, they do so in volumes which bear a simple whole number ratio to each other and to the volume of products, provided the products are also in gasesous state under similar conditions of temperature and pressure. This is what Gay Lussac’s law states. For example, in the reaction of carbon monoxide with oxygen, two volumes of carbon monoxide react with one volume of oxygen to give two volumes of carbon dioxide under similar conditions of temperature and pressure. 2CO + O₂ —> 2 CO₂ 2 vol. 1 vol. 2 vol. The volume ratio of carbon monoxide, oxygen and carbon dioxide is 2 : 1 : 2.

31.

How does Avogadro’s law explain Gay-Lussac’s law of gaseous volumes ?

Answer»

Avagadro’s law states that equal volumes of all gases contain equal number of molecules under similar conditions of temperature and presure. Since, when gases react chemically, they do so in volumes which bear a simple whole number ratio to each other and to the volume of products, provided the products are also in gasesous state under similar conditions of temperature and pressure. This is what Gay Lussac’s law states. For example, in the reaction of carbon monoxide with oxygen, two volumes of carbon monoxide react with one volume of oxygen to give two volumes of carbon dioxide under similar conditions of temperature and pressure.

2CO + O₂ —> 2 CO₂

2 vol. 1 vol. 2 vol.

The volume ratio of carbon monoxide, oxygen and carbon dioxide is 2 : 1 : 2.

Discussion

32.	What is the empirical formula of octane (C8H18 )?
Answer» M.F. of octane = C₈H₁₈ = (C₄H₉)₂ ∴ E.F. of octane = C₄H₉

32.

What is the empirical formula of octane (C8H18 )?

Answer»

M.F. of octane = C₈H₁₈ = (C₄H₉)₂

∴ E.F. of octane = C₄H₉

Discussion

33.	A gas cylinder holds 85 g of a gas ‘X’. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure Calculate the molecular weight of ‘X’.
Answer» V.D. of 'X' = weight of certain volume of gas/weight of equal vol. of H₂ Under similar conditions of temp. and pressure ∴ V.D. of gas 'X' = wt of 1 gas cylinder of gas 'X'/ wt. of 1 gas cylinder of H₂= 85 g/8.5g V.D. of 'X' =10 ∴ Molecular weight of gas 'X' = 2 x 10 = 20 g

33.

A gas cylinder holds 85 g of a gas ‘X’. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure Calculate the molecular weight of ‘X’.

Answer»

V.D. of 'X' = weight of certain volume of gas/weight of equal vol. of H₂

Under similar conditions of temp. and pressure

∴ V.D. of gas 'X' = wt of 1 gas cylinder of gas 'X'/ wt. of 1 gas cylinder of H₂= 85 g/8.5g

V.D. of 'X' =10

∴ Molecular weight of gas 'X' = 2 x 10 = 20 g

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the following :Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]

A vessel contains X number of molecules of hydrogen gas at a certain temperature &amp; pressure. Under the same conditions of temperature &amp; pressure, how many molecules of nitrogen gas would be present in the same vessel.

Select the correct answer from A, B, C and D : The ratio between the number of molecules in 2g of hydrogen and 32g of oxygen is: k (A) 1 : 2 (B) 1 : 0.01 (C) 1:1 (D) 0.01 : 1 [H = 1, O = 16]

Explain the following :Is it possible to change the temperature and pressure of a fixed mass of gas without changing its volume ?

Explain the following :One mole of hydrogen contains 2 x 6.023 x 1023 atoms of hydrogen where as one mole of helium contains 6.023 x 1023 atoms of helium

A gas cylinder can hold 1 kg. of H2 at room temp. &amp; press: State the law that helped you to arrive at the above result.

what volume (in ml)of 0.2M H2SO4 solution should be mixed with the 40ml of 0.1M NaOH SOLUTIONsuchthat the resulting solution has concentration of H2SO4 as (6/55)M

A compound is formed by 24 grams of X and 64 grams of oxygen, if X = 12 and O = 16. Find the simplest formula of the compound. ”

What is the important information given by a balanced chemical equation of decomposition of hydrogen peroxide ?

What are the limitations of a chemical equation?

“When stating the volume of a gas, the pressure and temperatrue should be also given.” Why ?

The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.Which sample contains the maximum number of molecules. If the temp, and pressure of gas A are kept constant, then what will happen to the volume of A when the no. of molecules is doubled.

The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.If this ratio of gas vols. refers to reactants and products of reaction – gas law observed is ….

Determines the relation between molecular weight and vapour density.

What is ‘mole scale’ of a compound?

State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if its V.D. is 8.

State Avogadro’s Law. A cylinder contains 68g of ammonia gas at s.t.p. 1. What is the volume occupied by this gas? 2. How many moles and how many molecules ammonia are present in the cylinder? [N = 14, H = 1]

The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OWhat volume, at s.t.p., is occupied by the number of moles determined in (1) (i) ?

What is the volume (measured in dm3 or litres occupied by one mole of gas at S.T.P. ?

The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

The vapour density of carbon dioxide [C = 12, O = 16] is (A) 32 (B) 16 (C) 44 (D) 22

A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2g, hence the relative molecular mass of the gas is : (A) 5 (B) 10 (C) 15 (D) 20

Why relative atomic mass is compared with 1 /12 of carbon?

Why is the term relative atomic mass used for atomic mass of an element ?

Define the term atomic weight :

Mention the term defined by the following : The mass of a given volume of gas compared to the mass of an equal volume of hydrogen.

From A, B, C, D, which weighs the least — A : 2 g. atoms of Nitrogen B : 1 mole of Silver C : 22.4 litres of oxygen gas at 1 atmospheric pressure and 273K D : 6.02 x 1023 atoms of carbon. [Ag = 108, N=14, 0=16, C=12]

The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2OIf the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?

A compound having empirical formula X2 Y is made of two elements X and Y. Find its molecular formula. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density 25.

How does Avogadro’s law explain Gay-Lussac’s law of gaseous volumes ?

What is the empirical formula of octane (C8H18 )?

A gas cylinder holds 85 g of a gas ‘X’. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure Calculate the molecular weight of ‘X’.

A vessel contains X number of molecules of hydrogen gas at a certain temperature & pressure. Under the same conditions of temperature & pressure, how many molecules of nitrogen gas would be present in the same vessel.

A gas cylinder can hold 1 kg. of H2 at room temp. & press: State the law that helped you to arrive at the above result.