Table of contents  
Introduction  
Scientific Notation  
Significant Figures  
Precision and Accuracy  
Units & Dimensional Analysis  
Conversion of Units  
Density 
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Many times in the study of chemistry, one has to deal with experimental data as well as theoretical calculations.
There are meaningful ways to handle the numbers conveniently and present the data realistically with certainty to the extent possible like:
Example: We can write 232.508 as 2.32508 × 10^{2} in scientific notation. Similarly, 0.00016 can be written as 1.6 × 10^{–4}.
Thus, we can write 232.508 as 2.32508 × 10^{2} in scientific notation. Note that while writing it, the decimal had to be moved to the left by two places and the same is the exponent (2) of 10 in the scientific notation.
Similarly, 0.00016 can be written as 1.6 × 10^{–4}. Here the decimal has to be moved four places to the right and (– 4) is the exponent in the scientific notation.
Q.1. Which of the following options is not correct?
(a) 8008 = 8.008 x 10^{3}
(b) 208 = 3
(c) 5000 = 5.0 x 10^{3}
(d) 2.0034 = 4
Ans: (d)
Solution:
2.0034 = 4
Q.2. Exponential notation in which any number can be represented in the form, Nx 10^{n }here N is termed as
(a) non –digit term
(b) digit term
(c) numeral
(d) base term
Ans: (b)
Solution:
In exponential notation N × 10^{n}, N is a number called digit term which varies between 1.000 and 9.000….
(i) Addition And Subtraction of Significant Figures
The result cannot have more digits to the right of the decimal point than either of the original numbers.
12.11
18.0
1.012
31.122
Here, 18.0 has only one digit after the decimal point, and the result should be reported only up to one digit after the decimal point, which is 31.1.
(ii) Multiplication and Division of Significant Figures
The result must be reported in these operations with no more significant figures as in the measurement with the few significant figures.
2.5 × 1.25 = 3.125
Since 2.5 has two significant figures, the result should not have more than two significant figures. Thus, it is 3.1.
Q.1. How many significant figures in each term?
(a) 34.6209 = 6
(b) 0.003048 = 4
(c) 5010.0 = 5
(d) 4032.090 = 7
Q.2. Solve the following equations using the correct number of significant figures.
(a) 34.683 + 58.930 + 68.35112 = 161.964
(b) 45001  56.355  78.44 = 44866
(c) 0.003 + 3.5198 + 0.0118 = 3.535
(d) 36.01  0.4  15 = 21
Q.3. Solve the following equations using the correct number of significant figures.
(a) 98.1 × 0.03 = 3
(b) 57 × 7.368 = 4.2 × 10^{2}
(c) 8.578/4.33821 = 1.977
(d) 6.90/2.8952 = 2.38
Q.4. How many significant figures in each term?
(a) 1.40 × 10^{3} = 3
(b) 6.01 = 3
(c) 02947.1 = 5
(d) 583.02 = 5
Example: If the true value for a result is 2.00 g.
(a) Student ‘A’ takes two measurements and reports the results as 1.95 g and 1.93 g.
These values are precise as they are close to each other but are not accurate.
(b) Another student repeats the experiment and obtains 1.94 g and 2.05 g as the results for two measurements.
These observations are neither precise nor accurate.
(c) When a third student repeats these measurements and reports 2.01g and 1.99 g as the result.
These values are both precise and accurate.
Example: How unit of work / energy i.e. joule, in S.I. system is related with unit erg in C.G.S system?
Dimension of work = force x displacement = MLT^{2} × L = ML^{2}T^{2}
1 joule = 1 kg (1 metre)^{2} × (1sec)^{2}
⇒ 1 × Kg × 1000 gm/1Kg × [1 metre × 100 cm/1 metre] × [1 sec]^{2}
⇒ 100 gm × (100)^{2} × 1 em^{2} × (1 sec)^{2}
⇒ 1000 × 10000 × 1 gm × 1 cm^{2} × 1 sec^{1}
⇒ 1 joule = 10^{7} erg
Similarly, we can deduce other conversion factors for other quantities in different units by the dimensional analysis method.
Another interesting example is the conversion of litre – atmosphere to joule (the SI unit of energy) by multiplying with two successive unit factors.
Thus,
1 L atm × (10^{3} m^{3}/1L) × 101.325 Pa/1 atm
= 101. 325 Pa M^{3}
Knowing that Pa = N/m^{2},
we can write
101.325 Pa m^{3} = 101.325 (N/m^{2}) m^{3}
= 101.325 N m = 101.325 J
Hence, 1 L atm = 101.325 J
Some Volume Measuring Devices
Relationship between Density, Mass and Volume
There are three common scales to measure temperature:
1. °C (degree Celsius)
2. °F (degree Fahrenheit)
3. K (kelvin)
Here, K is the SI unit.
Conversion Formulas:
Note: Temperature below 0 °C (i.e. negative values) are possible in the Celsius scale but in the Kelvin scale, the negative temperature is not possible.
Temperature Kelvin scale
Example 1. What is the mass of 1 L of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g cm^{−3}?
Solution. We know the relationship, 1 L = 1000 cm^{3} and
Also, density = mass/volume
We can write, mass = (volume) (density)
Therefore, the mass of 1 L of mercury is equal to
The mass in kilograms can be calculated as
(Remember,
are conversion factors with which we have to multiply for getting our answer in appropriate units).
Example 2. How unit of velocity i.e. kilometer/hour is related to unit meter/second.
Solution. We know the relationship, velocity = total displacement/total time
Prefixes in S.I. system
Name  Symbol  Quantity 
yotta  Y  10^{24} 
zetta  Z  10^{21} 
exa  E  10^{18} 
peta  P  10^{15} 
tera  T  10^{12} 
giga  G  10^{9} 
mega  M  10^{6} 
kilo  k  10^{3} 
hecto  h  10^{2} 
deca  da  10 
deci  d  10^{1} 
centi  c  10^{2} 
miIIi  m  10^{3} 
micro  M  10^{6} 
nano  n  10^{9} 
pico  P  10^{12} 
fempto  f  10^{15} 
atto  a  10^{18} 
zepto  z  10^{21} 
yocto  y  10^{24} 
Solved Examples:
Q.1. In scientific notation,0.00016 can be written as
(a) 1.6 x 10^{4}
(b) 1.6 x 10^{3 }
(c) 1.6 x 10^{2}
(d) 1.6 x 10^{1}
Ans: (a)
Solution:
0.00016 can be written as 1.6 × 10^{4} in scientific notation
Q.2. Addition of 6.65 x 10^{4} and 8.95 x 10^{3}, in terms of scientific notation will be
(a) 7.545×10^{4 }
(b) 75.45×10^{3 }
(c) 754.5×10^{2 }
(d) 75.45×10^{0}
Ans: (a)
Solution:
6.65 × 10^{4} + 8.95 × 10^{3}
= (6.65 + 0.895) × 10^{4} = 7.545 × 10^{4}
Q. 3. The substraction of two numbers
2.5 x 10^{2} 4.8 x 10^{ 3 }gives the following value.
(a) 2.02 x 10^{3 }
(b) 2.02 x 10^{2}
(c) 2.02 x 10^{1}
(d) 2.02 x 10^{0}
Ans: (b)
Solution:
2.5 × 10^{2}  4.8 × 10^{3}
= 2.5 × 10^{2} — (0.48 × 10^{2}) = 2.02 × 10^{2}
Q.4. A refers to the closeness of various measurements for the same quantity. B is the agreement of a particular value to the true value of the result. A and B respective are
(a) A → Significant figures, B → accuracy
(b) A → accuracy, B → precision
(c) A → Precision, B → accuracy
(d) A → significant figures, → precision
Ans: (c)
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