Q13.68 - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a)  $\mathbf{3}\mathbf{.}\mathbf{10}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{g}$ of 0.125 m ethylene glycol (${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}$ ) from ethylene glycol and water. 

(b) 1.20 kg of 2.20 mass %  ${\mathbf{HNO}}_{\mathbf{3}}$ from 52.0 mass %  ${\mathbf{HNO}}_{\mathbf{3}}$.

Step-by-Step Solution

Answer

1. The mass of water required to prepare the given solution is 39.9 kg.
2. Consider 50.7g of 52.0 % (mass) solution and it should be diluted to 1.20 kg solution.

Step 1: Concentration of the Solution

Solution can be formed from the dissolution of the solute in the solvent. The solute decides the nature of the solution.

Molality can be defined as the ratio of the mass of the solute to the mass of the solution present in kilogram measure.

$\mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{ } \mathbf{of} \mathbf{solute}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solvent} \mathbf{(} \mathbf{in} \mathbf{kilogram} \mathbf{)}}$

Number of moles can be defined as the ratio of the mass of the atom/molecule and molar mass of the atom/molecule.

$\mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar}\mathbf{ } \mathbf{Mass}}$

Density can be defined as the ratio of the mass of the matter to the volume of the matter.

$\mathbf{Density}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}}$

Step 2:Subpart (a)

Mass of Ethylene glycol,  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{10}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{g}$

Molar Mass of Ethylene glycol,  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{67}\mathbf{g}\mathbf{/}\mathbf{mol}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{10}\mathbf{\times }{\mathbf{10}}^{\mathbf{2}}\mathbf{g}}{\mathbf{62}\mathbf{.}\mathbf{07}\mathbf{ }\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{99}\mathbf{ }\mathbf{mol} \end{gathered}$$

Number of moles of Ethylene glycol, ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{99}\mathbf{mol}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number}\mathbf{ }\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}}{\mathbf{Mass}\mathbf{ }\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{ }\mathbf{Solvent}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{kg}\mathbf{ }\mathbf{)}} \\ \mathbf{0}\mathbf{.}\mathbf{125}\mathbf{m}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{99}\mathbf{ }\mathbf{mol}}{\mathbf{Mass}\mathbf{ }\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{ }\mathbf{Solvent}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{Kg}\mathbf{ }\mathbf{)}} \\ \mathbf{Mass}\mathbf{ }\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{ }\mathbf{Solvent}\mathbf{=}\mathbf{39}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{kg} \end{gathered}$$

Therefore, the mass of the solvent is 39.9 kg.

Step 3: Subpart (b)

The final concentration of  $HN{O}_3$ is 2.20% (mass).

The final mass of the solution is 1.20 kg

The initial concentration of $HN{O}_3$ is 52.0%(mass)

The mass of solution taken can be calculated by using the formula:

 $C_1\times M_1=C_2\times M_2$--------- (1)

Where, $C_1$  is the initial concentration, $M_1$  is the initial mass, $C_2$  is the final concentration, $M_2$  is the final mass of solution.

Substitute the given values in (1)

 $$\begin{gathered} C_1\times M_1=C_2\times M_2 \\ 52.0\%\times M_1=2.20\%\times 1.20 kg \\ {M}_1=\frac{2.20\%\times 1.20 kg}{52.0\%} \\ =0.0507 kg \\ =50.7 g \end{gathered}$$

Hence, consider 50.7g of 52.0 % (mass) solution and it should be diluted to 1.20 kg solution.