1. Basic Concepts Of Chemistry

Question 1

Significant figures

Answer 1

Significant figures are the digits within a number to determine how accurate the data value is .

Question 2

Rules to determine Significant values.

Answer 2

• All non zero number are significant.
• Zeros between any two non zero number are significant.
• Zeros after the decimal are significant.
• Zero and coefficient of scientific notations are significant.
• Any leading zeros are not significant.
• Zeros in larger numbers are not significant.

Question 3

Significant values rules for arithmetic operations.

Answer 3

Addition and Subtraction
Look for least amount of decimal places .

Multiplication and Division
Look for least amount of significant figures.

Question 4

Precision and Accuracy

Answer 4

Precision
The closeness of measurement for a particular quantity.

Accuracy
The particular value closest to the true value of a quantity.

Question 5

Dimensional analysis
1. Length

Answer 5

-1in = 2•54cm
-1ft = 12in
-1yd = 3ft
-1m = 1•0936yd
-1mi = 1609•3m
-1mi = 1•609km

Question 6

Dimensional analysis
2. Mass

Answer 6

-1kg = 2•205lb
-1lb = 453•6g
-1oz = 28•3495g

Question 7

Dimensional analysis
3.Volume

Answer 7

-1L = 1dm³
-1ml = 1cm³
-1dm = 10cm
-1L = 1•057qt(quart)
-1ft³ = 28•317L
-1gal = 4qt

Question 8

Dimensional analysis
4. Time

Answer 8

-1hr = 60min
-1min = 60sec
-1hr = 3600sec

Question 9

Law of Conservation of Mass

Answer 9

Mass can neither be created nor be destroyed.

In all physical and chemical changes there is no net change in mass during the process.

Question 10

Law of Definite Proportion

Answer 10

Irrespective of the source, a given compound contains same element combined together in same proportion by mass.

A given compound always contains exactly the same proportion of elements by weight ,whether the element source is organic or inorganic the proportion remains same .

Question 11

Law of Multiple Proportion

Answer 11

When two elements combine to form more than one compound then masses of the element combined with a fixed mass of other element are in the ratio of small whole numbers.

Eg.-
H² +O = H²O
2g 16g 18g

   H² + O² = H²O²
  2g    32g    34g

Here Oxygen (16g and 32g) combine with fix mass of Hydrogen (2g) bears a ratio ( 16:32 i.e 1:2 )

Question 12

Gay Lussac’s Law of Gaseous Volume

Also known as Law of Definite Proportion by Volume

Answer 12

When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are in same temperature.

Question 13

Avogadro’s Law

Answer 13

“Gases are equal volumes at same temperature and pressure contains same amount of molecules.”

Question 14

Atomic mass

Answer 14

“The atomic mass unit is assigned a value exactly equal to the mass of 1/12th of ¹²C atom . ”

So, 1amu = 1.660×10^-24 g
Amu= atomic mass unit

Average Atomic mass
It is the atomic mass used for calculations . All the isotopes relative abundance and atomic mass are taken in accordance to measure the average mass .

**Average Atomic mass “= (relative abundance of isotope)(atomic mass of isotope)+………….

Question 15

Molecular mass and Formula mass

Answer 15

Molecular mass is the mass of any compound whole formula mass is of ions making compound.

Question 16

Mole and molar mass

Answer 16

**One mole = atoms in 12g of ¹²C or Carbon-12 **

One mole of any substance, is always same i.e

Na= 6.022×10^23 atoms/mole

Na is avogadro’s constant.

**The mass of one mole of any substance in grams is called it’s molar mass . **

Question 17

Percentage Composition

Answer 17

It is used to check the purity of an element in a compound.

Mass percent of an element = (Mass of element in compound÷
Molar mass of the compound)×100

Question 18

Empirical formula

Answer 18

**Empirical formula represents the ratio of atoms present in the elements in the compound in simple whole numbers **

Steps -
1. Convert mass percent into grams if given if grams given ignore
2. Find number of moles of each element
3. Divide the ratio of moles such that they are in simple whole number
4. Use the ratio to write empirical formula
5. Find the empirical formula mass
6. Divide the given molar mass with EFM
7. Multiply the result with empirical formula
8. This is the molecular formula.

Question 19

Mass per cent

Answer 19

As the name say mass per 100gm

Mass per cent = (mass of solute ÷ mass of solution)×100

Question 20

Mole Fraction

Answer 20

Ratio of no. Of moles of an element in compound

Say there are two elements whose nom of moles are Na and Nb then,

Mole fraction of A = _____Na______
(Na+Nb)

Mole fraction of B = _____Nb______
(Na+Nb)

Question 21

Molarity

Answer 21

The number of moles of solute dissolve in 1L of solution. Used to determine how concentrated the solution is.

Molarity= No. of moles of solute ÷
Volume of solution in L

Volume of solution= solute +solvent

Solute remains constant but water or any other solvent is added to it later and made together up to 1 L i.e the mass of solvent is changing with every other solute .

Molarity is affected by temperature since volume depends upon the temperature.

Question 22

Molality

Answer 22

The number of moles of solute dissolved in 1kg solvent.

Molality= No. of moles of solute ÷
Mass of solvent in kg

The mass of solute and solvent are fixed and mixed as it is .

Molality doesn’t depend upon the temperature since mass remains constant ( it never change to begin with)

Question 23

Limiting reactant

Answer 23

The reactant in a reaction who gets used up the fastest and produce the least amount of product is called Limiting reactant.

The other reactant in the reaction is excess reactant and the product produced by limiting reactant is called the theoretical yeild.