Class 11 : Scientific Notation, Significant Figures & Dimensional Analysis Class 11 Notes | EduRev
The document Scientific Notation, Significant Figures & Dimensional Analysis Class 11 Notes | EduRev is a part of the Class 11 Course Chemistry for JEE.
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:
Scientific Notation
Significant Figures
Dimension Analysis
Scientific Notation
Scientific Notation is a way of expressing numbers that are too big or too small to be conveniently written in decimal form.
In which any number can be represented in form N × 10^{n }(where n is an exponent having positive or negative values and N can vary between 1 to 10).
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.
Multiplication and Division for Exponential Numbers
These two operations follow the same rules which are there for exponential numbers, i.e.
Addition and Subtraction for Exponential Numbers
For these two operations, first, the numbers are written in such a way that they have the same exponent. After that, the coefficients (digit terms) are added or subtracted as the case may be.
Thus, for adding 6.65 × 10^{4} and 8.95 × 10^{3}, the exponent is made the same for both numbers. Thus, we get (6.65 × 10^{4}) + (0.895 × 10^{4}).
Examples:
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….
Significant Figures
The reliability of measurement is indicated by the number of digits used to represent it.
To express the measurement more accurately, we express it with digits that are known with certainty. These are called significant figures.
They contain all the certain digits plus one doubtful digit in a number.
Rules for Determining the Number of Significant Figures
All non-zero digits are significant. Example: - 6.9 has two significant figures. - 2.16 has three significant figures.
The decimal place does not determine the number of significant figures.
A zero becomes significant in case it comes in between non-zero numbers. Example: - 2.003 has four significant figures. - 4.02 has three significant figures.
Zeros at the beginning of a number are not significant. Example: - 0.002 has one significant figure. - 0.0045 has two significant figures.
All zeros placed to the right of a number are significant. Example: - 16.0 has three significant figures. - 16.00 has four significant figures.
Zeros at the end of a number without a decimal point are ambiguous.
In exponential notations, the numerical portion represents the number of significant figures. Example:0.00045 is expressed as 4.5 × 10^{-4} in terms of scientific notations. The number of significant figures in this number is 2, while in Avogadro's number (6.023 × 10^{23}) it is four.
The decimal point does not count towards the number of significant figures. Example: The number 345601 has six significant figures but can be written in different ways, as 345.601 or 0.345601 or 3.45601, all having the same number of significant figures.
Math with Significant Figures
(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.
Try yourself:Look at the addition of significant figures given below 12.11 + 18.0 + 1.012 = 31.122 The result reported in this addition should be
Explanation
The result cannot have moredigits to the right of the decimal point than either of the original numbers 12.110 + 18.000 + 1.012 = 31.122 Here, 18.0 has only one digit after the decimal point and the result should be reported only upto one digit after the decimal point which is 31.1
View Solution
Try yourself:The result reported in the following multiplication of significant figures, 2.5 ×1.25 = 3.125 should be
Explanation
In these operations, the result must reported with no more significant figures as are there 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
View Solution
Practice Questions
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
Precision and Accuracy
Every experimental measurement has some amount of uncertainty associated with it. However, one would always like the results to be precise and accurate. Precision and accuracy are often referred to while we talk about measurement.
Accuracy refers to the closeness of a measured value to a standard or known value. Example: If in the lab you obtain a weight measurement of 3.2 kg for a given substance, but the actual or known weight is 10 kg, then your measurement is not accurate. In this case, your measurement is not close to the known value.
Precision refers to the closeness of two or more measurements to each other. Using the example above, if you weigh a given substance five times, and get 3.2 kg each time, then your measurement is very precise. Precision is independent of accuracy.
You can be very precise but inaccurate, as described above. You can also be accurate but imprecise. Example: If on average, your measurements for a given substance are close to the known value, but the measurements are far from each other, then you have accuracy without precision.
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.
Units & Dimensional Analysis
Conversion of Units
The simplest way to carry out calculations that involve different units is to use dimensional analysis. The simplest way to carry out calculations that involve different units is to use dimensional analysis.
As there is a need to convert units from one system to the other. The method used to accomplish this is called the factor label method or unit factormethod or dimensional analysis.
In this method a quantity expressed in one unit is converted into an equivalent quantity with a different unit by using a conversion factor that expresses the relationship between units:
This is based on the fact that the ratio of each fundamental quantity in one unit with their equivalent quantity in another unit is equal to one For example, in the case of mass 1 kg = 2.205 pound = 1000 gm
In this way, any derived unit first expressed in dimension and each fundamental quantity like mass length time is converted in another system of the desired unit to work out the conversion factor.
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
Mass and Weight
Mass of a substance is the amount of matter present in it while weight is the force exerted by gravity on an object.
The mass of a substance is constant whereas its weight may vary from one place to another due to a change in gravity. The mass of a substance can be determined very accurately by using an analytical balance.
Volume
Volume has units of (length)^{3}. So, the volume has units of m^{3} or cm^{3} or dm^{3}. A common unit, litre (L) is not an SI unit, is used for the measurement of the volume of liquids. 1 L = 1000 mL, 1000 cm^{3} = 1 dm^{3}
Some Volume Measuring Devices
Density
The density of a substance is its amount of mass per unit volume. SI unit of density = SI unit of mass/SI unit of volume = kg/m^{3} or kg m^{–3}.
This unit is quite large and a chemist often expresses density in g cm^{–3}.
Relationship between Density, Mass and Volume
Temperature
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:
K = °C + 273.15
°F = (9/5) °C +32
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}
Try yourself:The temperature at which both the Celsius and Fahrenheit scales will have the same reading is:
Explanation
Suppose both read the same value as “x” Then as
4x = - 160 x = - 40º
View Solution
Try yourself:Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below: 1Pa = 1Nm^{-2} If the mass of air at sea level is 1034 g cm^{-2}, the pressure in pascal is:
Explanation
Pressure is the force (i.e., weight) acting per unit area & weight = mg
View Solution
Try yourself:The prefix femto stands for:
Explanation
1 femto = 10^{-15}
View Solution
Try yourself:N kg^{-1} is the unit of:
Explanation
Force = Mass * Acceleration Hence, a = F/m = N/kg^{-1} = N kg^{-1}
View Solution
Try yourself:Which of the following is correct?
Explanation
1 L = 1000 cm^{3 }= 1000 (0.1 dm)^{3} = 1dm^{3}
View Solution
Try yourself:The units, nanometer, Fermi, Angstrom and attometre, Arrangement in decreasing order are expressed as:
Explanation
1 nm = 10^{-9} m 1 fermi =10^{-15} m 1 A˚ = 10^{-10} m 1 attometre = 10^{-18} m
View Solution
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)
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!