Chemistry

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Payal Gupta

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1.50. Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as the standard. Since C-12 is used as the standard atom, one atomic mass unit is defined as one-twelfth of the mass of one carbon – 12 atoms. This is because it has an equal number of protons and neutrons (6) and makes up the majority of matter.

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Payal Gupta

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1.49. The empirical mass of ethene is half of its molecular mass. The empirical formula represents the simplest whole-number ratio of various atoms present in a compound.

Molecular Formula = n * Empirical formula

Empirical formula of Ethene = C2H4

Empirical Formula Mass = 14 amu= ½ Molecular Mass of Ethene

The ratio of Carbon and Hydrogen in the empirical formula is 1: 2.

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Payal Gupta

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1.48. 

Molarity

Molality

The molarity of a given solution is defined as the total number of moles of solute per litre of solution.

Molality is defined as the total moles of a solute contained in a kilogram of a solvent.

The mathematical expression is-

M = number of moles of the solute /Volume of solution given in terms of litres.

M = (g ? 1000)/(W ? V).

The mathematical expression is-

m = Numbers of moles of solute/Mass of solvent in kgs

m = (g ? 1000)/(W ? m).

Depends on the volume of the whole solution.

Depends on the mass of the solvent.

Unit sign expressed as (M).

Unit sign expressed as (m).

Molarity has a unit of mol/litre.

Molality has units of mol/kg.

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Payal Gupta

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1.47. These rules for determining the number of significant figures are stated below:

(1) All non-zero digits are significant. For example, in 285 cm, there are three significant figures and in 0.25 mL, there are two significant figures.

(2) Zeros preceding to first non-zero digit are not significant. Such zero indicates the position of decimal point. Thus, 0.03 has one significant figure and 0.0052 has two significant figures.

(3) Zeros between two non-zero digits are significant. Thus, 2.005 has four significant figures.

(4) Zeros at the end or right of a number are significant, provided they are on the right side of the decimal po

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Payal Gupta

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1.46.  (c) both a and b

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Payal Gupta

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1.45.  (d) 36 g of water

CH4 + 2O2 → CO2 + 2H2O

1 mol of CH4 reacts with 1 mol O2 of to produce 2 moles of H2O

It means that 16g of CH4 reacts with oxygen to produce (2x18)= 36g of water

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Payal Gupta

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1.44.  (b) Candela

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Payal Gupta

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1.43. (c) The mass of 1 mole of C-12 atoms = 12 g

1 mole of C- atoms = 6.022 * 1023 atoms

The mass of 1 atom of C-12 = 12 / (6.022 * 1023)

= 1.99 * 10–23 g

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Payal Gupta

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1.42.  (d) all the above

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