Nature of Matter, Scientific Notation and Units & Dimensional Analysis Class 11 Notes | EduRev
Chemistry for JEE
Class 11 : Nature of Matter, Scientific Notation and Units & Dimensional Analysis Class 11 Notes | EduRev
The document Nature of Matter, Scientific Notation and Units & Dimensional Analysis Class 11 Notes | EduRev is a part of the Class 11 Course Chemistry for JEE.
Anything that exhibits inertia is known as matter. The quantity of matter is its mass. Example: chalk table. In simple language, Anything which has mass and occupies space is called Matter. Matter can exist in three physical states viz. solid, liquid, and gas. The constituent particles of matter in these three states can be represented as shown in the figure: A Representation of the Solid, Liquid, and Gas States
In solids, these particles are held very close to each other in an orderly fashion and there is not much freedom of movement.
In liquids, the particles are close to each other but they can move around.
In gases, the particles are far apart as compared to those present in solid or liquid states and their movement is easy and fast.
Because of such an arrangement of particles, different states of matter exhibit the following characteristics:
Solids have definite volume and definite shape.
Liquids have a definite volume but not a definite shape. They take the shape of the container in which they are placed.
Gases have neither a definite volume nor a definite shape. They completely occupy the container in which they are placed.
Note: These three states of matter are interconvertible by changing the conditions of temperature and pressure.
Classification of Matter (At Macroscopic/Bulk Level)
This classification of matter is based upon the chemical composition of various substances. According to this matter can be further divided into two types, pure substance, and mixture. Mixtures are also of two types, homogenous mixtures and heterogeneous mixtures.
Classification of Matter
➢ Elements
The primary stuff present in all the substances is known as an element, whose smallest unit is known as an atom.
Sodium, copper, silver, hydrogen, oxygen, etc. are some examples of elements. They all contain atoms of one type. However, the atoms of different elements are different in nature.
Total 118 elements are known till the date of which 92 are naturally occurring elements rest are results of artificial transmutation. There are 88 metals, 18 nonmetals, and 6 metalloids.
➢ Compound
A non-elemental pure substance is called a compound in which more than one atom of elements is linked by chemical bonds formed due to chemical reaction.
The resulting molecule is an electrically neutral particle of constant continuous composition.
➢ Mixture
Mixtures are the aggregate of more than one type of pure substance whose chemical identity remains maintained even in mixtures. Their constituent ratio may vary, unlike compounds. Example: Sugar + Water = Sugar Syrup Gunpowder: 75% KNO_{3 }+ 10% Sulphur + 15% carbon
There are two types of mixture:
(a) Homogeneous (b) Heterogenous
Homogeneous mixtures are those whose composition for each part remains constant. Example: aqueous and gaseous solution.
Heterogeneous mixtures are those whose composition may vary for each and every part. Example: soil, concrete mixtures.
Table: Comparison between Homogeneous and Heterogeneous mixtures
Table: Comparison between Compound and Mixtures
Try Yourself!
Question 1:Which one of the following is not an element?
Explanation
Silica (SiO_{2)} is a compound and not an element.
Question 2:The most abundant element on the earth’s crust is:
Explanation
The most abundant element on earth’s crust is Oxygen.
Question 3:Which one of the following is not a mixture?
Explanation
Plaster of Paris is a compound (CaSO_{4}).
Question 4:Which one if the following is not a metalloid?
Explanation
Germanium is not a metalloid.
Scientific Notation, Precision & Accuracy
Uncertainty in Measurement
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).
Scientific Notation
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}, exponent is made same for both the numbers. Thus, we get (6.65 × 10^{4}) + (0.895 × 10^{4}).
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 the measurement.
Accuracy refers to the closeness of a measured value to a standard or known value. Example: If in 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.
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….
Q.3.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
4.Addition of 6.65 x 10^{4} and 8.95 x 10^{3}, in terms of scientific notation will be (a) 7.545 104 (b) 75.45 10 3 (c) 754.5 102 (d) 75.45 100 Ans: (a) Solution: 6.65 × 10^{4} + 8.95 × 10^{3} = (6.65 + 0.895) × 10^{4} = 7.545 × 10^{4}
Q.5.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.6.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)
Q.7.Which of the following statement is/are true? (a) Every experimental measurement has zero amount of uncertainty associated with it (b) One would always like the result to be precise and accurate (c) Precision and accuracy are often referred to while we talk about the measurement (d) Both (b) and (c) Ans: (d) Solution: Every experimental measurement has some amount of uncertainty associated with it.
Try Yourself!
Question 5:Which of the following statement is correct?
Explanation
Good accuracy means that the average value of measurements is close to the actual value. Different measurements may not be close to each other (which is a condition for precision).
Question 6:Two students X and Y report the weight of the same substance as 5.0g and 5.00g respectively. Which of the following statements is correct?
Explanation
5.00 is more accurate than 5.0 because the former has three significant figures while the latter has two.
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 which express the relationship between units:
This is based on the fact that ratio of each fundamental quantity in one unit with their equivalent quantity in other unit is equal to one For example, in 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 are converted in other system of 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 factor for other quantity in different unit by the dimensional analysis method. Another interesting example is the conversion of liter – 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 change in gravity. The mass of a substance can be determined very accurately by using an analytical balance.
➢ Volume
Volume has the units of (length)^{3}. So, 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 measurement of volume of liquids. 1 L = 1000 mL, 1000 cm^{3} = 1 dm^{3}
Some Volume Measuring Devices
➢ Density
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!
Question 7:The temperature at which both the Celsius and Fathrenheit scales will have the same reading is
Explanation
Suppose both read the same value as “x” Then as
or 4x = - 160 or x = -40º
Question 8:Pressure is determined as force per unit area of 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 But weight = mg
Question 9:The prefix femto stands for
Explanation
1 femto = 10^{-15}
Question 10:N kg^{-1} is the unit of
Explanation
Force = mass x acceleration. Hence, a = F/m = N/kg^{-1} = N kg^{-1}.
Question 11:Which of the following is correct?
Explanation
I L = 1000 cm^{3 }= 1000 (0.1 dm)^{3} = 1dm^{3}.
Question 12: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.
Question 13:Dimensions of pressure are same as that of
Explanation
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