Question

Na, so, is added in water to make 12% w/w
solution having density 1.2 g/mL. Correct statement
for resulting solution is/are
(a) Nearly 0.96 mole of Na, so, dissolved per kg
of H2O
(b) Normality of Na® ion is equals to normality of
sogion
(c) Molarity of solution is less than molality​

Answer 1

Answer: The correct statements are Option a and Option b

Explanation:

We are given:

12% (w/w) $Na_2SO_4$ solution. This means that 12 grams of $Na_2SO_4$ is present in 100 grams of solution

Mass of solvent = 100 - 12 = 88 g

To calculate the molality of $Na_2SO_4$, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$m_{solute}$ = Given mass of solute $(Na_2SO_4)$ = 12 g

$M_{solute}$ = Molar mass of solute $(Na_2SO_4)$ = 142 g/mol

$W_{solvent}$ = Mass of solvent = 88 g

Putting values in above equation, we get:

$\text{Molality of }Na_2SO_4=\frac{12\times 1000}{142\times 88}\\\\\text{Molality of }Na_2SO_4=0.96m$

This means that 0.96 moles of $Na_2SO_4$ is present in 1 kg of solvent that is water.

• Calculating the molarity of solution:

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.2 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.2g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100}{1.2}=83.33mL$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Mass of solute $(Na_2SO_4)$ = 12 g

Molar mass of $Na_2SO_4$ = 142 g/mol

Volume of solution = 83.33 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{12g\times 1000}{142g/mol\times 83.33mL}\\\\\text{Molarity of solution}=1.014M$

Thus, molarity of solution is greater than molality of solution.

• Calculating the normality of ions in the solution:

The chemical equation for the ionization of $Na_2SO_4$ follows:

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

1 mole of $Na_2SO_4$ produces 2 moles of sodium ions and 1 mole of sulfate ion.

Concentration of sodium ions = $(2\times 1.014)=2.028M$

Concentration of sulfate ions = $(1\times 1.014)=1.014M$

To calculate the normality of ion, we use the equation:

$N=nM$

where,

n = charge on ion

M = molarity of ions

Normality of sodium ion:

Charge on sodium ion = 1

$N_{Na^+}=1\times 2.028=2.028N$

Normality of sulfate ion:

Charge on sulfate ion = 2

$N_{SO_4^{2-}}=2\times 1.014=2.028N$

Thus, the normality of sodium ions is equal to the normality of sulfate ions.

Hence, the correct statements are Option a and Option b.