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**Conservation of Mass**According to this law during any chemical and physical change the total mass of product is equal to the total mass of reactants.

**Examples**

**9 g of KClO**_{3}heated produce 1.92 g of O_{2}and the residue KCl left behind weight 2.96 g show that the data illustrate the law of conservation of mass.

Solution:

\(\begin{matrix}

KClO_3& \xrightarrow{\Delta}& KCl& +& O_2& \\

4.9\;g& &2.96\;g&+&1.92\;g& \\

&&&=&\;4.88\;g&\\

\end{matrix}\)

**Law of Definite Proportion**According to this law, a pure chemical compound always contains same elements combined together in the same element combined together in the same definite proportion in the weight.

**Examples**

**16 g of Cu when treated with NO**_{3 }followed by ignition of the nitrate give two 2.90 g of CuO. In another experiment 1.15 of CuO upon reduction with H_{2}give 0.932 g of Cu.

Solution:

In 1^{st}case \(\frac{2.16}{2.70}\) \(\times\) 100 = 80%

% of Cu in CuO = 80 %

% of O in CuO = 20 %

\(\frac{0.92}{1.5}\) \(\times\) 100 = 20%

**Law of multiple proportion**According to this law when two elements combine to form two or more than two compounds, then the masses of one of the elements which combine with a fixed mass of other are in a simple whole no. ratio.

**Examples**

**Carbon is found to form oxides which contains 42.9 % and 27.3 % of a respectively shows the multiple proportion.**Solution:

= 100 - 42.9 = 57.1

= 100 – 27.3 = 72.7

1 part of O = 42.9 part of C

1 part of O = \(\frac{42.9}{37.1}\) = 0.75

In II^{nd}case

72.7 part of O = 27.3 part of C

1 part of O = \(\frac{27.3}{72.7}\) = 0.37

= \(\frac{0.75}{0.37}\):\(\frac{0.37}{0.37}\)

= 2 : 1

**Gay Lussac’s Law**

It states that whenever gases reacts with each other or combine with each other or combine with each other to form gases out product they always have a simple whole ratio by volume under similar condition of temperature and pressure.

\(\begin{matrix}

2H_2& +&O_2& \longrightarrow& 2H_2O&\\

2\;Vol&&1\;Vol& \longrightarrow& 2\;Vol&\\

2\;ml&&1\;ml& \longrightarrow& 2\;ml&\\

2\;l&&1\;l& \longrightarrow& 2\;l&\\

\end{matrix}\)

**Avogadro’s law **It states that equal no. of all gases contains equal no. of molecules under similar condition of temperature and pressure.

2H_2& +&Cl_2& \longrightarrow& 2HCl&\\

1\;Vol&&1\;Vol& \longrightarrow& 2\;Vol&\\

n\;molecules&&n\;molecules& \longrightarrow & 2n\;molecules&\\

1\;molecule&&1\;molecule& \longrightarrow & 1\;molecule&\\

\frac{1}{2}\;molecule&&\frac{1}{2}\;molecule& \longrightarrow & \frac{1}{2}\; molecule&\\

\end{matrix}\)

Molecules mass = 2 \(\times\) U.D

**Examples**

- Molar mass of 44 g = 1 mol

22400 ml at STP

1 mol = 22400 ml

5 mol = 22400 \(\times\) 1.5

**Atomic Mass**The atomic mass of an element is the no. of times on atom of that elements is heavier than 1 atom of C

**Examples**

**Calculate the no. of atom in 52 u of He.**Solution:

52 u of He

= 1 u = 4

= 13 \(\times\) N_{A}atoms

**Gram Atomic Mass**The atomic mass of an elements is express in g is known as gram atomic mass.

**Examples**

- 1 mol AlCl
_{3}= 1 mol Al^{3+}= 27g AL^{3+}= N_{A }\(\times\) Al^{3}ion

= 3 mol of Al^{-}= 106.5 g al^{-}= 3 N_{A}al^{-}ions

1 mol AlCl_{3}= total no. of ions = 4

1 mol AlCl_{3}= 4 N_{A }

**Examples**

**How many moles of CH**_{4}are required to produced 22 g CO_{2}after combustion.

Solution:

By formula

\(\begin{matrix}

CH_4&+&2H_2O&\rightarrow&CO_2&+&H_2O&\\

&1\;Mol&&&&2\;Mol&&\\

\\

CH_4&\rightarrow&CO_4&\\

22\;g&&144\;g&\\

\end{matrix}\)

1 g CO_{2 }= \(\frac{22}{44}\) = \(\frac{1}{2}\)

1 CO_{2 }= 1 CH_{4}\(\frac{1}{2}\)CO_{2}= \(\frac{1}{2}\)CH_{4}= 0.5 mole of CH_{4}22 g = \(\frac{1}{44}\) \(\times\) 22 = \(\frac{1}{2}\) CH_{4}= 0.5 mole of CH_{4}**Statement**\(\begin{matrix}

&CH_4& +& 2O_2& \longrightarrow& 2H_2O& +& CO_2& \\

&1\;Mol&&&&&&1\;Mol&&\\

&16\;g&&&&&&44\;g&&\\

&44\;g\;\;CO_2&&& \longrightarrow&&& 1\;Mol\;\;CH_4&\\

\end{matrix}\)

1 g = \(\frac{1}{44}\)

22 g = \(\frac{1}{44}\) \(\times\) 22 = \(\frac{1}{2}\)

**Limiting Reagent**Which completely consume in the reaction.

**Excess Reagent**Which do not completely consume in the reaction.

**Example**

**1 L of O**Solution:_{2}at STP is made to react 56.66 kg with 3 L of CO. Calculate the mass of each substance found after the reaction and also tell which is limiting reagent?\(\begin{matrix}

2CO&+&O_2&\longrightarrow& 2CO_2&\\

2\;Vol&&1\;Vol&&2\;Vol&\\

\end{matrix}\)

2 L \(\rightarrow\) 2 L

\(\begin{bmatrix}

2\;L\;\;O_2& \longrightarrow& O_2&\\

1\;g& \longrightarrow& \frac{1}{2}&\\

3\;L& \longrightarrow& \frac{1}{2}\;\times\;3\;=\;\frac{3}{2}&\\

\end{bmatrix}\)

Limiting Reagent O_{2}2CO + O_{2}\(\rightarrow\) 2CO_{2}4 l \(\rightarrow\) 28 g

1 l \(\rightarrow\) \(\frac{28}{22.4}\)

3l \(\rightarrow\) \(\frac{28}{22.4}\) \(\times\) 30 = \(\frac{840}{22.4}\) = 1.25 g

22.4 l \(\rightarrow\) 44 g CO_{2}1 l \(\rightarrow\) \(\frac{44}{22.4}\) = 1.964

2l \(\rightarrow\) 1.964 \(\times\) 2 = 3.929

**Solution**

It is a homogenous mixture of two or more compound present in any ratio.

**Concentration**Dilute

The concentration can be expressed as follows

**Mass % (w/w %)**

It is the mass of solute in grams present in 100 g of the solution.

% of component 1 = \(\frac{w_1}{w_1 + w_2}\) \(\times\) 100

% of component 2 = \(\frac{w_2}{w_1 + w_2}\) \(\times\) 100

10% NaOH solution (mass/mass) means that 10 g NaOH is dissolved in 100 g of the solution (90 g H_{2}O)**Volume present (v/v %)**

It is the vol. of salute in ml present in 100 ml of solution.

10 % ethyl alcohol solution means 10 ml ethyl alcohol is present in 100 ml of the salutation.**Mass/volume (w/v %)**

It is the mass salute in grams present in 100 ml of the solution. 10 percent NaOH solution (w/v) means that 10 g of NaOH is present in 100 ml of the solution.**Strength (g/l)**

It is the amount of solute in g present in 1 l of the solution.

1 g \(\rightarrow\) 100 ml of solution

1 g \(\rightarrow\) 100 ml of solution

**Example **

**If we have 100 ml of the solution contain 5 g solute what will be its strength.**

5 g \(\rightarrow\) 100 ml

50 g \(\rightarrow\) 1000 ml

2 l \(\rightarrow\) 10 g

1 l \(\rightarrow\) 5 g

**Molarity (Mol/L)**

It is defined as the no. of moles of solute dissolved per line of the solution. It is represented by capital (M)

Mathematically, M = \(\frac{No.\;of\;moles\;of\;solute}{vol.\;of \;solution(L)}\)

= \(\frac{given\;mass}{molar\;mass}\) \(\times\) \(\frac{1000}{V}\)

**Examples**

**Calculate the maturity of NaOH in the solution prepared by dissolving its four 4 g in enough water to form 250 ml of the solution.**Solution:

M = \(\frac{4}{40}\) \(\times\) \(\frac{100}{250}\) = 0.4 M

**Relation between strength and maturity**As we know

**Molality (m)**mol/kg

It is also equal to given mass \(\times\) 1000. (W

**Example**

**Commercial available concentrated Hal contain 38% by mass. What is the molality of this solution.**Solventing 1.19 (Density)

38% HCl by mass means that 38 ml of Hal is present in 100 g of the solution.

M = \(\frac{given\;mass}{molar\;mass}\) \(\times\) \(\frac{1000}{v}\)

Mass of solute = 38 g

Mass of solvent = 100 g

= \(\frac{38}{36.5}\) \(\times\) \(\frac{1000}{84.03}\) = 12.38

**Molarity Equation**m

**How many (m) grams BaCl**\(\frac{250}{100}\) \(\times\) \(\frac{1000}{1000}\) \(\times\) 20 m = 5.20 g_{2}are needed to prepare 100cm^{3}of 0.250 m. BaCl_{2}(Ba = 173 g).

= \(\frac{0.250}{544{\times}1000}\) \(\times\) \(\frac{1000}{100}\) = 5.20 g

**Normality (N)**g-equivalent/L

It is the no. of g - equivalent of the solute dissolved per liter of the solution.

For acid

\(\begin{bmatrix}

HCl,&Basicity\;=\;1\;=\;n&\\

H_2SO_4,&Basicity\;=\;2\;=\;n&\\

H_3PO_4,&Basicity\;=\;3\;=\;n&\\

\end{bmatrix}\)

For Base

For salts

**Examples**

**MgCl**_{2}= n = 2

mass = \(\frac{24+2(35.5}{2}\) = 47.5

AlCl_{2}= n= 3

Eq. mass = \(\frac{27 + 3(35.5)}{3}\) = 36.5

**Relation between molality and normality**N = \(\frac{givem\;mass}{eq.\;mass}\) \(\times\) \(\frac{1000}{u (ml)}\)

**Relation between normality and strength**N = \(\frac{given\;mass}{equ.\;mass}\) \(\times\) \(\frac{1}{v(L)}\)

**Normality Equation**N

**Examples**

**H**When 32.65 g of Zn react with Hal, 1mole of gas occupies 22.7 l at STP._{2}gas is prepared in the lab by reacting dal. Hal with 2n following reaction takes place. Calculate the vol of H_{2 }gas liberated at Z_{n }+ 2HCl \(\rightarrow\) 2ZnCl_{2 }+ H_{2}

Zn + 2HCl \(\rightarrow\) 2ZnCl_{2 }+ H_{2}65

65.3 g \(\rightarrow\) 22.7 l

1 g \(\rightarrow\) \(\frac{22.7}{65.3}\) L

32.65 g \(\rightarrow\) \(\frac{22.7}{65.3}\) \(\times\) 32.65 = 11.35

**PPM (Parts per Million (10 ^{6}**

It is no. of parts of solute by mass present in million parts by mass of the solution.

(PPM)

**Examples**

**Calculate the amount of C formed.**

2 A \(\rightarrow\) 4 B

1 A \(\rightarrow\) \(\frac{4}{2}\) \(\times\) 5

4 D \(\rightarrow\) 3 C

1 D \(\rightarrow\) \(\frac{3}{4}\) \(\times\) 6 = 4.5 C**Calculate the average atomic mass using the isotope**

% data

Molar Mass

H

99.985

1

^{2}H0.015

2

= \(\frac{1 \times 99.985}{100}\) = 0.99985

= \(\frac{2 \times 0.015}{100}\) = 0.0003

In the reaction A + B \(\rightarrow\) AB_{2}identify the limiting reagent if any in the following mixture. (1) 300 atoms of A + 200 molecules of B (2) 2 mole of A + 3 mole of B (3) 100 atoms of A + 100 molecules of B (4) 5 mole of A + 2.5 mole of B (5) 2.5 mole of A + 5 mole of B.

300 atoms of A \(\rightarrow\) 200 molecules of B

\(\rightarrow\) \(\frac{200}{300}\)= 0.6

Limiting regent = B

2 Mole of A \(\rightarrow\) 3 mole of B

Limiting reagent = A

**Limiting regent Nitrogen**28 g N \(\rightarrow\) 34 g NH

**Example**

**What will be the mass of C**_{12}

Mass of 6.022 \(\times\) 10^{23}atoms of carbon = 12 g

1 atom of carbon = \(\frac{12}{6.022{\times}10^{23}}\)

= 1.99 \(\times\) 10^{23}

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