Chapter 1: Some Basic Concepts of Chemistry

Ans: ${{H}_{2}}O$

The molecular mass of water, ${{\text{H}}_{\text{2}}}\text{O}$, can be calculated by the following steps given below:

= (2 $\times $ Atomic mass of hydrogen element) + (1 $\times $ Atomic mass of oxygen element)

= $\text{(2  }\!\!\times\!\!\text{  1}\text{.0084 u) + (1  }\!\!\times\!\!\text{  16}\text{.00 u)}$   

= 2.016 u + 16.00 u

= 18.016 u or 18.02 u

Ans:  $\text{C}{{\text{O}}_{\text{2}}}$

The molecular mass of carbon dioxide,$\text{C}{{\text{O}}_{\text{2}}}$, is calculated below:

= (1 $\times $ Atomic mass of carbon) + (2 $\times $ Atomic mass of oxygen)

= $\text{(1  }\!\!\times\!\!\text{  12}\text{.011 u) + (2  }\!\!\times\!\!\text{  16}\text{.00 u)}$   

= 12.011 u + 32.00 u

= 44.01 u

Ans: $\text{C}{{\text{H}}_{\text{4}}}$

The molecular mass of methane,$\text{C}{{\text{H}}_{\text{4}}}$, is calculated below in step by step manner:

= (1 $\times $ Atomic mass of carbon) + (4 $\times $ Atomic mass of hydrogen)

= $\text{(1  }\!\!\times\!\!\text{  12}\text{.011 u) + (4  }\!\!\times\!\!\text{  1}\text{.008 u)}$   

= 12.011 u +4.032 u

= 16.043 u

Ans: The given compound in the question is sodium sulphate and its formula is $\text{N}{{\text{a}}_{2}}S{{O}_{4}}$. Its molecular formula is calculated below: 

$\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=}\left[ \left( \text{2 }\!\!\times\!\!\text{ 23}\text{.0} \right)\text{+}\left( \text{32}\text{.066} \right)\text{+4}\left( \text{16}\text{.00} \right) \right]$

= 142.066 g

Now, let us find the mass percentage of each element in the given compound using the formula given below:

Mass percent of an element $=\dfrac{\text{Mass of that element in the compound }}{\text{Molar mass of the compound}}\times 100$

$\therefore $Mass percent of sodium:

$\text{=}\dfrac{\text{46}\text{.0 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 32.379

= 32.4%

Now, let us find the mass percentage of sulphur:

$\text{=}\dfrac{\text{32}\text{.066 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 22.57

= 22.6%

Now, the mass percentage of oxygen:

$\text{=}\dfrac{\text{64}\text{.0 g}}{\text{142}\text{.066 g}}\text{ }\!\!\times\!\!\text{ 100}$

= 45.049

= 45.05%

Ans: We are given that the percentage of iron by mass is 69.9% and the percentage of oxygen by mass is 30.1%.

Now, we can calculate the relative moles of iron by using the formula given below:

$=\dfrac{\%\text{ of iron by mass}}{\text{Atomic mass of iron}}$

$=\dfrac{69.9}{55.85}$

= 1.25

In the same way we can calculate the relative moles of oxygen:

$=\dfrac{\%\text{ of oxygen by mass}}{\text{Atomic mass of oxygen}}$

$=\dfrac{30.1}{16.00}$

= 1.88

Since we have relative moles of both elements, we can calculate the simpler molar ratio.

= 1.25: 1.88

= 1: 1.5

= 2: 3

So, now we can write the empirical formula of iron oxide as $F{{e}_{2}}{{O}_{3}}$.

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

Carbon dioxide is produced when one mole of carbon is burned in one mole of dioxygen (air) using the balancing equation.

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

There is only 16 g of dioxygen accessible, as stated in the question. In this case, it will react with 0.5 moles of carbon to produce 22 g of $C{{O}_{2}}$. In this sense, it is a limiter.

Ans: It is possible to express the balanced response of carbon combustion as follows:

$\text{C+}{{\text{O}}_{\text{2}}}\to \text{C}{{\text{O}}_{\text{2}}}$

It appears that just 16 g of dioxygen is accessible. In other words, it's a reaction-suppressing agent. This means that with just half as much carbon as dioxygen, 22 g of carbon dioxide may be produced.

Ans: 0.375 M aqueous Ans of sodium acetate means 1000 ml of Ans contains 0.375 moles of sodium acetate.

So, we can calculate the number of moles of sodium acetate in 500 ml. This is given below:

$=\dfrac{0.375}{1000}\times 500$

= 0.1875 mole

We are also given the molar mass of sodium acetate is $82.0245\text{ g mol}{{\text{e}}^{-1}}$. 

Required mass of sodium acetate can be calculated as:$=\left( 82.0245\text{ g mo}{{\text{l}}^{-1}} \right)\left( 0.1875\text{ mole} \right)$

= 15.38 g

Ans: We are given the mass percentage of nitric acid in the sample as 69%.

So, we can say that 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid $\text{(HN}{{\text{O}}_{\text{3}}}\text{)}$ is calculated as below: 

=  $\text{1 + 14 + (3  }\!\!\times\!\!\text{  16)}$

= 1 + 14 + 48

= $63\text{ g mo}{{\text{l}}^{-1}}$

As we have the molar mass of nitric acid so, we can find the number of moles in 69 g of nitric acid as given below:

$=\dfrac{69\text{ g}}{63\text{ g mo}{{\text{l}}^{-1}}}$

= 1.095 mol

Volume of 100 g of nitric acid And can be calculated as:

$=\dfrac{\text{Mass of solution}}{\text{density of solution}}$

$=\dfrac{100\text{ g}}{1.41\text{ g m}{{\text{L}}^{-1}}}$

$\text{=70}\text{.92 mL}$

$\text{=70}\text{.92 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ L}$

Now, we can calculate the concentration of nitric acid as:

$\text{=}\dfrac{\text{1}\text{.095 mole}}{\text{70}\text{.92 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{ L}}$

= 15.44 mol/ L

Therefore, the concentration of nitric acid is 15.44 mol/ L

Ans: In copper sulfate, we can see that there is one atom of copper so, we can say that 1 mole of $CuS{{O}_{4}}$ will have 1 mole of copper.

The molar mass of copper sulphate is calculated below:

$CuS{{O}_{4}}$ = $\text{63}\text{.5 + 32 + (4  }\!\!\times\!\!\text{  16)}$

= 63.5 + 32.0 + 64.0

= 159.5 g

We can say that 159.5 g of $CuS{{O}_{4}}$ will have 63.5 g of copper.

$\Rightarrow 100\text{ g of CuS}{{\text{O}}_{4}}$ will contain $\dfrac{\text{63}\text{.5 }\!\!\times\!\!\text{ 100 g}}{\text{159}\text{.5}}$ of copper.

So, the amount of copper that can be obtained from 100 g of $\text{CuS}{{\text{O}}_{\text{4}}}$$=$$\dfrac{\text{63}\text{.5 }\!\!\times\!\!\text{ 100 g}}{\text{159}\text{.5}}$

= 39.81 g

Ans: We are given the mass percentage of iron (Fe) as 69.9% and the percentage of oxygen (O) as 30.1%.

We know the atomic mass of iron is 55.85.

The number of moles of iron present in the oxide will be $=\dfrac{69.90}{55.85}$

$=1.25$

The atomic mass of oxygen is 16.

The number of moles of oxygen present in the oxide will be $=\dfrac{30.1}{16.0}$

$=1.88$

The ratio of iron to the oxygen in the oxide is calculated below:

= 1.25: 1.88

$=\dfrac{1.25}{1.25}:\text{ }\dfrac{1.88}{1.25}$

= 1: 1.5

= 2: 3

According to the ratio the formula of the oxide will be $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$.

Empirical formula mass of $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\text{ =  }\!\![\!\!\text{ (2  }\!\!\times\!\!\text{  55}\text{.85) + (3  }\!\!\times\!\!\text{  16}\text{.00) }\!\!]\!\!\text{  = 159}\text{.7}$ g/ mol

Given molar mass of$\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ = 159.69 g/ mol

Multiplying the empirical formula by n gives the molecular formula for the molecule. This is calculated below:

$\text{n = }\dfrac{\text{Molar mass}}{\text{Empirical formula mass}}\text{=}\dfrac{\text{159}\text{.69 g}}{\text{159}\text{.7 g}}$

= 0.999 = 1

So, the empirical formula of the oxide is $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ and the value of n is 1. Therefore, the molecular formula of the oxide is $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$.

Ans:The average atomic mass of chlorine is calculated below:

$\text{=}\left[ \left( \text{Fractional abundance of}{{\text{ }}^{\text{35}}}\text{Cl} \right)\left( \text{Molar mass of}{{\text{ }}^{\text{35}}}\text{Cl} \right)\text{+}\left( \text{Fractional abundance of}{{\text{ }}^{\text{37}}}\text{Cl} \right)\left( \text{Molar mass of }{{\text{ }}^{\text{37}}}\text{Cl} \right) \right]$$\text{=}\left[ \left\{ \left( \dfrac{\text{75}\text{.77}}{\text{100}} \right)\left( \text{34}\text{.9689u} \right) \right\}\text{+}\left\{ \left( \dfrac{\text{24}\text{.23}}{\text{100}} \right)\left( \text{36}\text{.9659u} \right) \right\} \right]$

= 26.4959 + 8.9568

= 35.4527 u

So, the average atomic mass of chlorine is 35.4527 u.

Ans: 2 Moles of carbon atoms are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of moles of carbon atoms in 3 moles of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

= $\text{2  }\!\!\times\!\!\text{  3 = 6}$

Ans: 6 moles of hydrogen atoms are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of moles of carbon atoms in 3 moles of${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

= $\text{3  }\!\!\times\!\!\text{  6 = 18}$

Ans: $6.023\text{ }\times \text{ 1}{{\text{0}}^{23}}$ molecules of ethane are present in each mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

Therefore, number of molecules in 3 moles of${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$.

$=3\times 6.023\times {{10}^{23}}=18.069\times {{10}^{23}}$