Chapter 1: Some Basic Concepts of Chemistry

Ans: First we have to find the least precise number to find the significant figures.

Ans: $\dfrac{\text{0}\text{.2856 }\!\!\times\!\!\text{ 298}\text{.15 }\!\!\times\!\!\text{ 0}\text{.112}}{\text{0}\text{.5785}}$

Least precise number of calculation = 0.112

Therefore, number of significant figures in the answer

Number of significant figures in the least precise number = 3

Ans: $\text{5 }\!\!\times\!\!\text{ 5}\text{.364}$

Least precise number of calculations = 5.364

Therefore, number of significant figures in the answer will be

Number of significant figures in 5.364 = 4

Ans: $\text{0}\text{.0125+0}\text{.7864+0}\text{.0215}$

Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

Ans: Molar mass of argon is calculated step by step.

$\text{=}\left[ \left( \text{35}\text{.96755 }\!\!\times\!\!\text{ }\dfrac{\text{0}\text{.337}}{\text{100}} \right)\text{+}\left( \text{37}\text{.96272 }\!\!\times\!\!\text{ }\dfrac{\text{0}\text{.063}}{\text{100}} \right)\text{+}\left( \text{39}\text{.9624 }\!\!\times\!\!\text{ }\dfrac{\text{90}\text{.60}}{\text{100}} \right) \right]\text{g mo}{{\text{l}}^{\text{-1}}}$

$\text{=}\left[ \text{0}\text{.121+0}\text{.024+39}\text{.802} \right]\text{g mo}{{\text{l}}^{\text{-1}}}$

So, the molar mass of argon is 39.947 g/ mol.

Ans: All the options are given in different forms so, we have to convert it into a number of atoms. These are given below:

Ans: $\text{1 mole of Ar = 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Ar}$

Therefore,

$\text{52 mole of Ar = 52 }\!\!\times\!\!\text{ 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of Ar}$

$\text{=3}\text{.131 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{25}}}\text{ atoms of Ar}$

Ans: $\text{1 atom of He = 4u of the He}$

Or this can be written as, 4 u of He = 1 atom of He

1. u of He = $\dfrac{\text{1}}{\text{4}}$ atom of He

1. of He = $\dfrac{\text{52}}{\text{4}}$ atom of He

= 13 atoms of He.

Ans: $\text{4g of He = 6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ atoms of He}$

Therefore, we can write:    

$\text{52g of He = }\dfrac{\text{6}\text{.022 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{ }\!\!\times\!\!\text{ 52}}{\text{4}}\text{ atoms of He}$

$\text{= 7}\text{.8286 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{24}}}\text{ atoms of He}$

Ans: 1 mole (44 g) of $\text{C}{{\text{O}}_{\text{2}}}$ will have 12 g carbon.

So, 3.38 g of $\text{C}{{\text{O}}_{\text{2}}}$ will have carbon = $\dfrac{\text{12g}}{\text{44g}}\text{ }\times \text{ 3}\text{.38g}$

= 0.9217 g

18 g of water will have 2 g of hydrogen.

So, 0.690 g of water contain hydrogen = $\dfrac{\text{2g}}{\text{18g}}\text{ }\times \text{ 0}\text{.690}$

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

=0.9984 g

So, the percentage of Carbon in the compound = $\dfrac{0.9217}{0.9984}\text{ }\times \text{ 100}$= 92.32%

Now, percentage of Hydrogen in the compound = $\dfrac{0.0767}{0.9984}\text{ }\times \text{ 100}$= 7.68%

Moles of carbon in the compound = $\dfrac{92.32}{12}=7.69$

Moles of hydrogen in the compound = $\dfrac{7.68}{1}=7.68$

Since, we have the number of moles of both the elements, the ratio of carbon to hydrogen will be:

7.69: 7.68 = 1: 1 

This is in the whole number, so the empirical formula will be CH.

Ans: We are given,

Weight of 10.0 L of the gas (at S.T.P) = 11.6 g

$\text{Weight of 22}\text{.4L of gas at STP = }\dfrac{\text{11}\text{.6g}}{\text{10}\text{.0L}}\text{ }\!\!\times\!\!\text{ 22}\text{.4L}$

$=25.984g$

$\approx 26g$

Hence, the molar mass of the gas is 26 g.

Ans: Empirical formula mass of CH = 12 + 1 = 13

Now, we can find the value of n. This is given below:

$\text{n = }\dfrac{\text{Molar mass of gas}}{\text{Empirical formula mass of gas}}$

$\text{=}\dfrac{\text{26g}}{\text{13g}}$

n = 2

Therefore,

$\text{Molecular formula of gas=}{{\left( \text{CH} \right)}_{\text{n}}}$

$\text{= }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$

Ans: We are given the molarity as 0.75 M which means in 1 L of Ans there are 0.75 moles of HCl. This can be converted into grams as:

$\text{=}\left[ \left( \text{0}\text{.75 mol} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{36}\text{.5gmo}{{\text{l}}^{\text{-1}}} \right) \right]\text{HCl is present in 1L of  water}$

$\text{=27}\text{.375g of HCl is present in 1L of water}$

Thus, 1000 mL of Ans contains 27.375 g of HCl.

So, the amount of HCl present in 25 mL of Ans

$\text{=}\dfrac{\text{27}\text{.375g}}{\text{1000mL}}\text{  }\!\!\times\!\!\text{  25mL}$

= 0.6844 g

The chemical reaction given in the question is:

$\text{CaC}{{\text{O}}_{\text{3}\left( \text{s} \right)}}\text{+2HC}{{\text{l}}_{\left( \text{aq} \right)}}\to \text{CaC}{{\text{l}}_{\text{2}\left( \text{aq} \right)}}\text{+C}{{\text{O}}_{\text{2}\left( \text{g} \right)}}\text{+}{{\text{H}}_{\text{2}}}{{\text{O}}_{\left( \text{l} \right)}}$

$\text{2 mol of HCl }\left( \text{2 }\!\!\times\!\!\text{ 36}\text{.5=73g} \right)$reacts with $\text{1 mol of CaC}{{\text{O}}_{\text{3}}}\left( \text{100g} \right)$ .

So, the amount of$\text{CaC}{{\text{O}}_{\text{3}}}$ that will react with $\text{0}\text{.6844g = }\dfrac{\text{100}}{\text{71}}\text{ }\!\!\times\!\!\text{ 0}\text{.6844g}$

= 0.9639 g

Ans: 1 mol of $\text{Mn}{{\text{O}}_{\text{2}}}$ reacts with 4 mol of HCl.

The molar mass of $\text{Mn}{{\text{O}}_{\text{2}}}$is 87g and the molar mass of HCl is 36.5 g. For 4 moles of HCl, the mass will be 146 g.

So, for 5.0 g of $\text{Mn}{{\text{O}}_{\text{2}}}$ will react with:

$=\dfrac{146}{87}\text{ }\times \text{ 5 = 8}\text{.4 g}$

Therefore, 8.4 g of HCl will completely react with 5 g of $\text{Mn}{{\text{O}}_{\text{2}}}$.