Chapter 1: Some Basic Concepts of Chemistry

Ans: The masses of dioxygen that will combine with the fixed dinitrogen mass are 32 g, 64 g, 32 g, and 80 g, respectively. The mass ratios of dioxygen are 1: 2: 1: 5. In light of this fact, the supplied experimental data is in accordance with the law of multiple proportions. One element's mass multiplied by the fixed mass of another element must be small whole numbers if two elements combine to produce more than one compound, according to this rule.

Ans: All the conversions are given below:

Ans: $\text{1 km = 1 km  }\!\!\times\!\!\text{  }\dfrac{\text{1000 m}}{\text{1 km}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{100 cm}}{\text{1 m}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{10 mm}}{\text{1 cm}}$

$\text{1km = 1}{{\text{0}}^{\text{6}}}\text{mm}$

$\text{1 km = 1 km  }\!\!\times\!\!\text{  }\dfrac{\text{1000 m}}{\text{1 km}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 pm}}{\text{1}{{\text{0}}^{\text{-12}}}\text{ m}}$

$\text{1km=1}{{\text{0}}^{\text{15}}}\text{pm}$

$\text{Hence,  1 km = 1}{{\text{0}}^{\text{6}}}\text{ mm = 1}{{\text{0}}^{\text{15}}}\text{ pm}$

Ans: $\text{1 mg = 1 mg  }\!\!\times\!\!\text{  }\dfrac{\text{1 g}}{\text{1000 mg}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 kg}}{\text{1000 g}}$

$\text{1mg = 1}{{\text{0}}^{\text{6}}}\text{ kg}$

$\text{Hence, 1mg = 1}{{\text{0}}^{\text{-6}}}\text{ kg = 1}{{\text{0}}^{\text{6}}}\text{ ng}$

iii) 1 mL = ……………. L ……………… $\text{d}{{\text{m}}^{\text{3}}}$

Ans: $\text{1 mL = 1 mL  }\!\!\times\!\!\text{  }\dfrac{\text{1 L}}{\text{1000 mL}}$

$\text{1 mg = 1}{{\text{0}}^{\text{-3}}}\text{ L}$

$\text{1 mL = 1 c}{{\text{m}}^{\text{3}}}\text{ = 1 c}{{\text{m}}^{\text{3}}}\text{  }\!\!\times\!\!\text{  }\dfrac{\text{1 dm }\!\!\times\!\!\text{ 1 dm }\!\!\times\!\!\text{ 1 dm}}{\text{10 cm }\!\!\times\!\!\text{ 10 cm }\!\!\times\!\!\text{ 10 cm}}$

$\text{1 mL = 1}{{\text{0}}^{\text{-3}}}\text{ d}{{\text{m}}^{\text{3}}}$

$\text{Hence, 1 mL = 1}{{\text{0}}^{\text{-3}}}\text{ L = 1}{{\text{0}}^{\text{-3}}}\text{ d}{{\text{m}}^{\text{3}}}$

Ans: From the question we can see that the time taken to cover the distance is 2.00ns. This can be written as:

$\text{=2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}\text{s}$

We know the speed of light$\text{= 3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{ m}{{\text{s}}^{\text{-1}}}$

So, the distance travelled by light in 2.00ns will be:

$=\text{Speed of light }\times \text{ Time taken}$

$\text{=}\left( \text{3}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{m}{{\text{s}}^{\text{-1}}} \right)\left( \text{2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}\text{s} \right)$

$\text{=6}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{m}$

= 0.600 m

Ans: Reactants that serve as limiting agents restrict the amount of a response. An initial reaction product is consumed before any more products are created, which stops the reaction and limits the quantity of products that are formed.

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 200 molecules of B will react with 200 atoms of A, leaving 100 atoms of A unusable as a result. B is, thus, the limiting reagent in the reaction.

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. In other words, 2 moles of A will react with just 2 moles of B. Consequently, 1 mol of A will not be used up in the process. This means that A is considered to be the limiting reagent.

Ans: According to the given reaction, 1 atom of A reacts with 1 molecule of B. The mixture is stoichiometric where no limiting reagent is present.

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of B may be combined with 2.5 mol of A. There will be 2.5 mol of a remaining. B is, thus, the limiting reagent in this reaction.

Ans: It is shown that 1 atom of A interacts with 1 molecule of B in the given reaction. Because of this, only 2.5 mol of A may be combined with 2.5 mol of B. There will be 2.5 mol of a remaining. A is, thus, the limiting reagent in this reaction.

Ans: First, we have to balance the given chemical equation.

${{\text{N}}_{\text{2}\left( \text{g} \right)}}\text{+3}{{\text{H}}_{\text{2}\left( \text{g} \right)}}\to \text{2N}{{\text{H}}_{\text{3}\left( \text{g} \right)}}$

From this equation we can say, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g)of ammonia.

=$\text{2}\text{.00 }\times \text{1}{{\text{0}}^{\text{3}}}\text{ g}$of nitrogen will react with$\dfrac{\text{6 g}}{\text{28 g}}\text{ }\!\!\times\!\!\text{ 2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ g}$ dihydrogen.

=$\text{2}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g}$ of dinitrogen will react with 428.6 g of dihydrogen.

We are given the amount of dihydrogen = $\text{1}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ g}$

Therefore,${{\text{N}}_{\text{2}}}$is the limiting reagent.

We can say that 28 g of ${{\text{N}}_{2}}$ will produce 34 g of $\text{N}{{\text{H}}_{3}}$. 

Hence, mass of ammonia produced by $\text{2000 g of  }{{\text{N}}_{\text{2}}}\text{ =}\dfrac{\text{34 g}}{\text{28 g}}\text{  }\!\!\times\!\!\text{  2000 g}$

= 2428.57 g

Ans: ${{\text{N}}_{2}}$ is the limiting reagent while${{\text{H}}_{\text{2}}}$ is the excess reagent. So, hydrogen will be left unreacted.

Ans: Mass of dihydrogen left unreacted can be calculated as 

$\text{=1}\text{.00 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{g - 428}\text{.6g}$

= 571.4 g

Ans: The molar mass of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ is given below:

$\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ = $\text{(2  }\!\!\times\!\!\text{  23) + 12}\text{.00 + (3  }\!\!\times\!\!\text{  16)}$

= 106 g/mol

So, 1 mole of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ means 106 g of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$.

Therefore, for 0.5 mol of $\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$ can be calculated as:

$\text{0}\text{.5 mol of  N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ = }\dfrac{\text{106 g}}{\text{1 mole}}\text{  }\!\!\times\!\!\text{  0}\text{.5 mol N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

$\text{=53g N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

$\text{= 0}\text{.50 M N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{=0}\text{.50mol/L N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$

Hence,$\text{0}\text{.50 mol N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$is present in 1 L of water or $\text{53g N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}$is present in 1 L of water.

Ans: Let us first write the reaction between dihydrogen and dioxygen. The reaction will be:

$\text{2}{{\text{H}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{O}}_{\text{2}\left( \text{g} \right)}}\to \text{2}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( \text{g} \right)}}$

Dioxygen reacts with two volumes of dihydrogen to generate two volumes of water vapour.

As a result, ten volumes of dihydrogen will react with five volumes of dioxygen to generate ten volumes of water Vapours.

Ans:To convert the given numbers into basic units we have to convert it into meters and kilograms. All the numbers are converted into meters and kilograms, and are written below:

Ans: $\text{1pm=1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{28}\text{.7pm=28}\text{.7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{=2}\text{.87 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-11}}}\text{m}$

Ans: $\text{1pm=1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{15}\text{.15pm=15}\text{.15 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

$\text{=1}\text{.515 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}\text{m}$

Ans: $\text{1mg = 1}{{\text{0}}^{\text{-3}}}\text{g}$

$\text{25365mg=2}\text{.53657 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{g}$

Since,

$\text{1g=1}{{\text{0}}^{\text{-3}}}\text{kg}$

$\text{2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{1}}}\text{g=2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-1}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{kg}$

$\text{25365mg=2}\text{.5365 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{kg}$$$

Ans: As we can see that all the atoms are given in grams so, by counting the number of atoms in each element we can find the answer. These are solved step by step and written below:

Ans: $\text{1g of Au (s)=}\dfrac{\text{1}}{\text{197}}\text{mol of  Au (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{197}}\text{ atoms of  Au (s)}$

$\text{=3}\text{.06 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of  Au (s)}$

Ans: $\text{1g of  Na (s)=}\dfrac{\text{1}}{\text{23}}\text{mol of  Na (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{23}}\text{ atoms of  Na (s)}$

$\text{=0}\text{.262 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of  Na (s)}$

$\text{=26}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of  Na (s)}$

Ans: $\text{1g of Li (s)=}\dfrac{\text{1}}{\text{7}}\text{mol of  Li (s)}$

$\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{7}}\text{ atoms of  Li (s)}$

$\text{=0}\text{.86 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of  Li (s)}$

$\text{=86}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of  Li (s)}$

Ans: $\text{1g of  C}{{\text{l}}_{\text{2}}}\text{ (g)=}\dfrac{\text{1}}{\text{71}}\text{mol of  C}{{\text{l}}_{\text{2}}}\text{ (g)}$

Molar mass of $\text{C}{{\text{l}}_{\text{2}}}$ is 71 (2 $\times $ 35.5)

\[\text{=}\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}{\text{71}}\text{ atoms of  C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

\[\text{=0}\text{.0848 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of  C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

\[\text{=8}\text{.48 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{21}}}\text{ atoms of  C}{{\text{l}}_{\text{2}}}\text{ (g)}\]

Hence, 1 g of Li (s) has the largest number of atoms.

Ans: First we have to find the mole fraction of ethanol.

Mole fraction of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH=}\dfrac{\text{Number of moles }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}{\text{Number of moles of solution}}$

$\text{0}\text{.40=}\dfrac{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}}{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+}{{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}}\text{                   }.................\text{(1)}$

Number of moles present in 1 L water:

${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=}\dfrac{\text{1000 g}}{\text{18 g mo}{{\text{l}}^{\text{-1}}}}$

${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{=55}\text{.55mol}$

Substituting the value of ${{\text{n}}_{{{\text{H}}_{\text{2}}}\text{O}}}$in equation (1),

$\dfrac{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}}{{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+55}\text{.55}}\text{=0}\text{.040}$

${{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=0}\text{.040}{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{+}\left( \text{0}\text{.040} \right)\left( \text{55}\text{.55} \right)$

$\text{0}\text{.96}{{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=2}\text{.222mole}$

${{\text{n}}_{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}}}\text{=2}\text{.314mole}$

$\text{Molarity of solution =}\dfrac{\text{2}\text{.314 mol}}{\text{1 L}}$

So, the molarity of the Ans is 2.314 M.

Ans: We know that 1 mole of carbon means 12 gram of carbon is there and in terms of number of atoms, there are $\text{6}\text{.023 }\times \text{ 1}{{\text{0}}^{\text{23}}}$ atoms in 1mole of carbon. This can be written as:

$1\text{ mole = 12 g = 6}\text{.023 }\times \text{ 1}{{\text{0}}^{23}}$

So, mass of one$^{\text{12}}\text{C}$ atom$\text{=}\dfrac{\text{12g}}{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}}$

$\text{=1}\text{.993 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-23}}}\text{g}$