Question

10g of $$N{H_4}Cl$$ (molar mass 53.5g/mol) when dissolved in 1000g water lowered the freezing point by $${0.637^o}C$$. The degree of hydrolysis of the salt is $$x \times {10^{ - 6}}$$ , if the degree of dissociation is 0.75. The molal depression constant of water is $$1.86kgmo{l^{ - 1}}K$$. identify x/9(nearest integer).

Answer 1

Hint: The hydrolysis of salt is referred to as the neutralization reaction in which acid and bases are reacted together for the formation of salt. Many salts ionize in water to form acids and bases and get dissociated into respective ions existing as a hydrate ion in aqueous solution. The salts which are produced by reacting a strong base and strong acid are neutral in nature. The salts formed by the reaction of strong base and weak acid is basic in nature.

Complete answer:
The molar concentration is defined as the ratio of the number of moles to the volume of solution. So the number of moles of $$N{H_4}Cl$$ is $$\text{number of moles} = \dfrac{{\text{gram weight}}}{{\text{molecular mass}}}$$
Gram weight= 10g
Molecular mass= 53.5g/mol
Substituting the values in formula
$$number of moles = \dfrac{{10}}{{53.5}}$$=0.1869 moles
So now we know that 1000g of water= 1L of water
Now the molar concentration is
$$molar concentration = \dfrac{{\text{number of moles}}}{{\text{volume of solution}}}$$
Number of moles= 0.1869moles
Volume of solution = 1L
Substituting the values in the formula we get,
$$molar concentration = \dfrac{{0.1869}}{1}$$=0.1869M
The relationship between the hydrolysis constant, base dissociation constant and the ionic product is the following
$$\text{hydrolysis constant} ({K_H}) = \dfrac{{\text{ionic product of water}({K_w})}}{{\text{base dissociation constant}({K_B})}}$$
Ionic product of water= $$1 \times {10^{ - 14}}$$
Base dissociation constant =$$1.8 \times {10^{ - 5}}$$
Substituting the values in the formula we get:
$$hydrolysis constant({K_H}) = \dfrac{{1 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}} = 5.5 \times {10^{ - 10}}$$
The degree of hydrolysis is
$$h = \sqrt {\dfrac{{{K_h}}}{C}} = \sqrt {\dfrac{{5.5 \times {{10}^{ - 10}}}}{{0.189}}} = 5.4 \times {10^{ - 5}}$$=$$54 \times {10^{ - 6}}$$
Here h= degree of hydrolysis
$${K_h}$$ is hydrolysis constant and
C Is the molar concentration
The degree of hydrolysis is
$$54 \times {10^{ - 6}}$$=$$x \times {10^{ - 6}}$$
So x=54

So $$\dfrac{x}{9} = \dfrac{{54}}{9} = 6$$

Note: Ammonium chloride is odourless and a white solid. It is soluble in water, ethanol and methanol. The molal depression constant is defined as the freezing point in the solution in which 1g of solute is dissolved in 100g of solvent. The molal depression in freezing point is also known as cryoscopic constant.