Mole Concept

Volumetric analysis is based on the principle of equivalence in that substances react in the ratio of their equivalents. At the equivalence point of the reaction, involving the reactants A and B,
Number of gram equivalents of A $$=$$ Number of gram equivalents B.
If $$V_A$$ ml of solution A having normality $$N_A$$ react just completely with $$V_B$$ ml of solution B having normality $$N_B$$, then
$$\dfrac{N_AV_A }{1000}=\dfrac{N_BV_B}{1000}\implies N_AV_A=N_BV_B$$      ...... $$(1)$$
Equation $$(1)$$ called normality equation is very useful in numerical calculations of volumetric analysis. Basically, the equivalent mass of a substance is defined as the parts by mass of it, which combine with or displace $$1.0078$$ parts ($$\approx 1\; part$$) by mass of hydrogen, $$8$$ parts by mass of oxygen and $$35.5$$ parts by mass of chlorine. Mass of a substance expressed in gram equal to its equivalent mass is called gram equivalent mass. Equivalent mass of a substance is not constant, but depends upon the reaction in which the substance participates. Equivalent mass of an acid in acid-base reaction is its mass in gram which contains $$1$$ mole of replaceable $$H^+$$ ions $$(=1.0078\; g\; \approx 1\; g)$$. On the other hand, equivalent mass of a base is its mass which contains $$1$$ mole of replaceable $$OH^-$$ ions. $$1\; g$$ equivalent mass each of an acid and base on reaction gives salt and $$1$$ mole of water $$(=18\; g)$$. Equivalent mass of an oxidising agent is its mass, which gains $$1$$ mole of electrons. It can be obtained by dividing the molecular mass or formula mass by the total decrease in oxidation number of one or more elements per molecule. On the other hand, equivalent mass of a reducing agent is the mass of the substance which loses $$1$$ mole of electrons. It can be calculated by dividing the molecular or formula mass of the substance by the total increase in oxidation number of one or more elements per molecular or formula mass.
Water is added to $$3.52$$ g of $$UF_{6}$$. The products formed are $$3.08$$ g of a solid (containing only $$U, O$$ and $$F$$) and only $$0.8$$ g of a gas. The gas (containing fluorine and hydrogen only) contains $$95$$ percent by mass fluorine. Assume that the empirical formula is same as molecular formula. ($$U=238, F=19$$).

Water is said to be soft water if it produces sufficient foam with the soap and water that does not produce foam with soap is known as hard water. Hardness has been classified into two types (i) Temporary hardness (ii) Permanent hardness.

Temporary hardness is due to presence of calcium and magnesium bicarbonate. It is simply removed by boiling as shown below:

$$Ca(HCO_{3})_{2} \xrightarrow{\triangle } CaCO_{3} \downarrow +CO_{2} \uparrow+H_{2}O$$

$$Mg(HCO_{3})_{2} \xrightarrow{\triangle } MgCO_{3} \downarrow +CO_{2} \uparrow+H_{2}O$$

Temporary hardness can also be removed by addition of slaked lime, $$Ca(OH)_{2}$$.

$$Ca(HCO_{3})_{2}+Ca(OH)_{2} \rightarrow 2CaCO_{3} \downarrow +2H_{2}O$$

Permanent hardness is due to presence of sulphate and chlorides of $$Ca, Mg$$ etc. It is removed by washing soda as shown below:

$$CaCl_{2}+Na_{2}CO_{3} \rightarrow CaCO_{3} \downarrow +2NaCl$$

$$CaSO_{4}+Na_{2}CO_{3} \rightarrow CaCO_{3} \downarrow +Na_{2}SO_{4}$$

Permanent hardness also removed by ion exchange resin process as given below:

$$2RH+Ca^{2+} \rightarrow R_{2}Ca+2H^{+}$$

$$2ROH+SO_{4}^{2-} \rightarrow R_{2}SO_{4}+2OH^{-}$$

The degree of hardness of water is measured in terms of ppm of $$CaCO_{3}$$ is present in $$10^{6}\:g$$ of $$H_{2}O$$. If any water contain $$120$$ ppm of $$MgSO_{4}$$, its hardness in terms of $$CaCO_{3}=100$$ ppm.

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride $$(KCl)$$ and potassium sulphate $$(K_2SO_4)$$. Potash production is often reported as the potassium oxide$$(K_2O)$$ equivalent or the amount of $$K_2O$$ that could be made from a given mineral. $$KCl$$ costs $$50$$ per $$kg$$.
Samples of $$O_2, CO $$ and $$ CO_2$$ under the same conditions of temperature and pressure contain the same number molecules represented by $$X$$. The molecules of oxygen occupy V liters and have a mass of 8 g under the same conditions of temperature and pressure.