Chapter 1: Some Basic Concepts of Chemistry

Ans: Molarity (M) of the Ans can be calculated by the formula given below:

$=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$

This can be further written as

$=\dfrac{\text{Mass of sugar / molar mass of sugar}}{\text{2 L}}$

Putting the values, we get:

$\text{=}\dfrac{\text{20 g/}\left[ \left( \text{12 }\!\!\times\!\!\text{ 12} \right)\text{+}\left( \text{1 }\!\!\times\!\!\text{ 22} \right)\text{+}\left( \text{11 }\!\!\times\!\!\text{ 16} \right) \right]\text{ g}}{\text{2 L}}$

$\text{=}\dfrac{\text{20 g/ 342 g}}{\text{2 L}}$

$\text{=}\dfrac{\text{0}\text{.0585mol}}{\text{2L}}$

$\text{=0}\text{.02925 mol }{{\text{L}}^{\text{-1}}}$

So, the molar concentration of sugar is $\text{=0}\text{.02925 mol }{{\text{L}}^{\text{-1}}}$.

Ans: The molar mass of methanol is calculated below: $\left( \text{C}{{\text{H}}_{3}}\text{OH} \right)=\left( 1\times 12 \right)+\left( 4\times 1 \right)+\left( 1\times 16 \right)$

$\text{=32 g mo}{{\text{l}}^{\text{-1}}}$

$\text{=0}\text{.032 kg mo}{{\text{l}}^{\text{-1}}}$

Molarity of the methanol will be $\text{=}\dfrac{\text{0}\text{.793 kg }{{\text{L}}^{\text{-1}}}}{\text{0}\text{.032 kg mo}{{\text{l}}^{\text{-1}}}}$

$=24.78\text{ mol }{{\text{L}}^{-1}}$

(Since density is mass per unit volume)

Now, to find the volume we have to write the formula,

${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{=}{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$

1 is for the given Ans and 2 is for the Ans to be prepared.

Putting the values, we get:

$\left( 24.78\text{ mol }{{\text{L}}^{-1}} \right){{\text{V}}_{1}}=\left( 2.5\text{L} \right)\left( 0.25\text{ mol }{{\text{L}}^{-1}} \right)$

${{\text{V}}_{1}}=0.0252\text{ L}$

${{\text{V}}_{1}}=25.22\text{ mL}$

Ans: Force per unit area of a surface is described as pressure. This is calculated below:

$\text{P=}\dfrac{\text{F}}{\text{A}}$

$\text{=}\dfrac{\text{1034g }\!\!\times\!\!\text{ 9}\text{.8 m}{{\text{s}}^{\text{-2}}}}{\text{c}{{\text{m}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1 kg}}{\text{1000 g}}\text{ }\!\!\times\!\!\text{ }\dfrac{{{\left( \text{100} \right)}^{\text{2}}}\text{ c}{{\text{m}}^{\text{2}}}}{\text{1}{{\text{m}}^{\text{2}}}}$

$=1.01332\times {{10}^{5}}\text{ kg }{{\text{m}}^{-1}}\text{ }{{\text{s}}^{-2}}$

We know the relation of force can be written as:

$\text{1 N=1 kg m }{{\text{s}}^{\text{-2}}}$

Then we can write,

$\text{1 Pa=1 N}{{\text{m}}^{\text{-2}}}\text{=1 kg }{{\text{m}}^{\text{-2}}}\text{ }{{\text{s}}^{\text{-2}}}$

$\text{1 Pa=1 kg }{{\text{m}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-2}}}$

$\text{Pressure=1}\text{.01332 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{ Pa}$

Ans: The kilogram is the SI unit of mass in the SI system (kg). An international kilogram prototype is defined as one kilogram.

Ans: The multiples are matched with their prefixes in the table given below:

Ans: Significant figures are those digits that have meaning and are recognized to be assured of their value.

They are used to show uncertainty in an experiment or a computed number. 

For example, if 18.2 mL is the result of an experiment, then 18 is certain while 2 is uncertain, and the total number of significant figures is 3.

Since the final digit signifies uncertainty, significant figures are defined as the sum of all of a number's decimal places.

Ans: 1 part out of 1 million $\left( \text{1}{{\text{0}}^{\text{6}}} \right)$parts is equal to 1 ppm.

Therefore, mass percent of $15$ ppm chloroform in water will be

$\text{=}\dfrac{\text{15}}{\text{1}{{\text{0}}^{\text{6}}}}\text{ }\!\!\times\!\!\text{ 100}$

$\simeq 1.5\times {{10}^{-3}}%$

Ans: 100 g of the sample will have $1.5\text{ }\times \text{ 1}{{\text{0}}^{-3}}$of $\text{CHC}{{\text{l}}_{\text{3}}}$.

So 1000 g of the sample contains will have $1.5\text{ }\times \text{ 1}{{\text{0}}^{-2}}$of $\text{CHC}{{\text{l}}_{\text{3}}}$ .

Therefore, molality of chloroform in water is calculated as:

$\text{=}\dfrac{\text{1}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{ g}}{\text{Molar mass of CHC}{{\text{l}}_{\text{3}}}}$

For we want molar mass of chloroform.

$\text{Molar mass of CHC}{{\text{l}}_{\text{3}}}\text{=12}\text{.00+1}\text{.00+3}\left( \text{35}\text{.5} \right)\text{ = 119}\text{.5 g mo}{{\text{l}}^{\text{-1}}}$

$\therefore $ Molality of chloroform in water$\text{=0}\text{.0125 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{m}$

$\text{=1}\text{.25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{m}$

Ans: If we want to express the numbers in scientific notation, we must first put the number in decimal form and multiply it by 10 with some power. These are given below:

1. 0.0048

Ans: $0.0048=4.8\times {{10}^{-3}}$

1. 234,000

Ans: $234,000=2.34\times {{10}^{5}}$

1. 8008

Ans: $8008=8.008\times {{10}^{3}}$

1. 500.0

Ans: $500.0=5.000\times {{10}^{2}}$

1. 6.0012

Ans: $6.0012=6.0012\times {{10}^{0}}$

Ans: There are some rules to find the number of significant figures and by following the rules the significant figures are given below:

1. 0.0025

Ans: There are 2 significant figures.

1. 208

Ans: There are 3 significant figures.

1. 5005

Ans: There are 4 significant figures.

1. 126,000

Ans: There are 3 significant figures.

1. 500.0

Ans: There are 4 significant figures.

1. 2.0034

Ans: There are 5 significant figures.

Ans: These can be written as:

1. 34.216

Ans: 34.216 = 34.2

1. 10.4107

Ans: 10.4107 = 10.4

1. 0.04597

Ans: 0.04597 = 0.0460

1. 2808

Ans: 2808 = 2810