Some Basic Concepts of Chemistry

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

(Natural abundance of 1H x molar mass ) + (Natural abundance of 2H x molar mass of 2H)

Natural abundance of  1H = 99.985

Natural abundance of 2H = 0.015

Average atomic mass = 99.985 + 0.030 100

100.015 100 = 1.00015u

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

(a) Is this statement true?  

Ans: Yes the given statement is true

 

(b) If yes, according to which law?

Ans: Multiple law of proportions: According to the Law of Multiple proportions, when two elements combine to generate more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of tiny whole numbers.

 

(c) Give one example related to this law

Ans:

C (g) + O (g) -> CO (g)

12 g    16 g       28 g

C (g) + O2 (g) -> CO2 (g)

12 g     32 g &n

...more

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

If all gases are at the same temperature and pressure, Gay lussac's law of gaseous volumes states that gases combine or are created in a chemical reaction in a simple volume ratio.

H2 (g) +          Cl2 (g) →           2HCL (g)

1 volume      1 volume        2 volume

22.4 litre       22.4 litre        44.8 litre

2N2 (g) +      O2

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Molecular mass of Ca3 (PO4)2 = 310.18 g / mole 

Given mass of calcium 4 x 30= 120 g

Given mass of phosphorous =  31 x 2 =62 g

Given mass of oxygen = 16 x 8 = 128 g

Mass percent of Calcium = 120 310.18 x 100 = 38.71%

Mass percent of Phosphorous = 62 310.18  x 100 = 20%

Mass percent of Oxygen = 128 310.18 x 100 = 41.29 %

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

The number of moles of a substance (known as the solute) dissolved in precisely 1 litce of a solution is known as molarity (solvent and solute combined). As a result, the formula for estimating molarity is as follows:

Molarity = n u m b e r o f m o l e s o f s o l u t e v o l u m e i n l i t r e

The term molarity is also used to refer to molar concentration. As a result, molar concentration measurement is based on the volume of liquid in which a substance is dissolved. It's vital to remember that the volume is in litres, so if we have volume in mL we need to convert that in liters.

Molality is the number of moles of substance (also k

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

SI unit of the mole is mol. The amount of a substance that contains as many particles or entities as there are atoms in exactly

12 g (0.012 kg) of the C-12  isotope is defined as a mole. One mole is defined as follows:

1 mole =  6.023 x 1023

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

On solving the above equation, the result is 5.4. All non-zeroes digits are significant. The significant figure is 2.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Mass of 1 mole of C-12 = 12g

1 mole contains 6.022*1023 atoms.

Thus, mass of 6.022*1023 atoms=12g

 Mass of 1 atom of carbon =126.022*1023 g
                                         =1.99*10−23 g

Thus, mass of one atom of C-12 is 1.99*10−23 g

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

In AB, 2 g of A combines with 5 g of B.

So, 4 g of A combines with 10 g of B.

In AB2, 2g of B combines with 10 g of B, So 4g of A combines with 20 g of B.

In A2B3 , 4G of B combines with 5 g of B.

In A2B3 4 g of B combined with 15 g of B.

So, the ratio between different masses of B which combine with fixed mass (4g) of A is 10: 20: 5: 15, that is, 2: 4: 1: 3.

Hence, the ratio is simple. Therefore, the law of multiple proportions is applicable.

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

According to the law of multiple proportions, when two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other is always in the ratio of tiny numbers.

 Example:

1. Compounds of carbon and oxygen:

C and O react to form two different compounds CO and CO2. In CO, 12 parts by mass of C reacts with 16 parts by mass of 0 .

In CO2 ,12 parts by mass of C reacts with 32 parts by mass of O .

If the mass of C is fixed at 12 parts of mass then the ratio in the masses of oxyg

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