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Some Basic Concepts of Chemistry - Properties of Matter and their Measurement

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Matter is defined as anything that possesses mass and occupies space. It exists in three physical states: solid, liquid, and gas, which are interconvertible by changing conditions of temperature and pressure.

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Physical properties are those which can be measured or observed without changing the identity or the composition of the substance, such as color, odor, melting point, and boiling point.

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The International System of Units (SI) defines seven base units: length (meter, mm), mass (kilogram, kgkg), time (second, ss), electric current (ampere, AA), thermodynamic temperature (kelvin, KK), amount of substance (mole, molmol), and luminous intensity (candela, cdcd).

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Mass is the amount of matter present in a substance and is constant, while weight is the force exerted by gravity on an object and can vary with location.

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Density is the amount of mass per unit volume. Its SI unit is kg⋅m−3kg \cdot m^{-3}, but it is frequently expressed in g⋅cm−3g \cdot cm^{-3} in laboratory work.

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Temperature is measured using three common scales: Celsius (∘C^\circ C), Fahrenheit (∘F^\circ F), and Kelvin (KK). The Kelvin scale is the SI unit and does not use the degree symbol.

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Scientific notation is used to express very large or very small numbers in the form N×10nN \times 10^n, where NN is a number between 1.000...1.000... and 9.999...9.999... and nn is an exponent.

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Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain. They reflect the precision of a measurement.

📐Formulae

T(K)=t(∘C)+273.15T(K) = t(^\circ C) + 273.15

∘F=95(∘C)+32^\circ F = \frac{9}{5}(^\circ C) + 32

Density=MassVolumeDensity = \frac{Mass}{Volume}

1 L=1000 mL=1000 cm3=1 dm31\ L = 1000\ mL = 1000\ cm^3 = 1\ dm^3

1 m3=106 cm31\ m^3 = 10^6\ cm^3

💡Examples

Problem 1:

Convert the average human body temperature, 37∘C37^\circ C, into Fahrenheit and Kelvin scales.

Solution:

To convert to Fahrenheit: ∘F=95(37)+32=66.6+32=98.6∘F^\circ F = \frac{9}{5}(37) + 32 = 66.6 + 32 = 98.6^\circ F To convert to Kelvin: K=37+273.15=310.15 KK = 37 + 273.15 = 310.15\ K

Explanation:

We use the standard conversion formulae for temperature scales to relate Celsius to Fahrenheit and Kelvin.

Problem 2:

A substance has a mass of 5.74 g5.74\ g and occupies a volume of 1.2 cm31.2\ cm^3. Calculate its density while considering significant figures.

Solution:

Density=5.74 g1.2 cm3=4.78333... g⋅cm−3Density = \frac{5.74\ g}{1.2\ cm^3} = 4.78333...\ g \cdot cm^{-3} Rounding to two significant figures (since 1.21.2 has only two): Density=4.8 g⋅cm−3Density = 4.8\ g \cdot cm^{-3}

Explanation:

Density is mass divided by volume. In multiplication and division, the result must be reported with the same number of significant figures as the measurement with the least number of significant figures.

Problem 3:

Perform the addition of 12.1112.11, 18.018.0, and 1.0121.012 and report the sum with correct significant figures.

Solution:

12.1118.00+1.01231.122\begin{array}{r} 12.11 \\ 18.0 \phantom{0} \\ + 1.012 \\ \hline 31.122 \end{array} Rounded result: 31.131.1

Explanation:

In addition, the result cannot have more digits to the right of the decimal point than any of the original numbers. Since 18.018.0 has only one digit after the decimal, the answer is rounded to 31.131.1.