Q13.69 P - Chemistry: Molecular Nature Of Matter And Change

Question

How would you prepare the following aqueous solutions? 

(a) 1.50 kg of 0.0355 m ethanol (${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}$ ) from ethanol and water 

(b) 445 g of 13.0 mass % HCl from 34.1 mass % HCl.

Step-by-Step Solution

Answer

1. The mass of ethanol required is 2.45 g.
2. Consider 169.6 g of 34.1 % (mass) solution and it should be diluted to 445 g with water to get 13.0%(mass) 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.

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

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

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

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 Ethanol,  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{g}$

Molar Mass of Ethanol, ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{46}\mathbf{g}\mathbf{/}\mathbf{mole}$

Molality of Ethanol, ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0355}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles}}{\mathbf{Mass} \mathbf{of} \mathbf{the} \mathbf{Solvent} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{0}\mathbf{.}\mathbf{0355}\mathbf{m}\mathbf{=}\frac{\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}}}{\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{Kg}} \end{gathered}$$

$$\begin{gathered} \mathbf{0}\mathbf{.}\mathbf{0355}\mathbf{m}\mathbf{=}\frac{\frac{\mathbf{Mass}}{\mathbf{46}\mathbf{g}\mathbf{/}\mathbf{mole}}}{\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{Kg}} \\ \mathbf{0}\mathbf{.}\mathbf{0355}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{46}\mathbf{g}\mathbf{/}\mathbf{mole}} \\ \mathbf{0}\mathbf{.}\mathbf{0355}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }\mathbf{46}\mathbf{=}\mathbf{Mass} \\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Ethanol}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{45}\mathbf{g} \end{gathered}$$

Hence, the mass of the ethanol solute required is 2.45g.

Step 3: Subpart (b)

The final concentration of HCl is 13.0% (mass).

 

The final mass of the solution is 445 g

 

The initial concentration of HCl is 34.1%(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)

 

 $$\begin{gathered} C_1\times M_1=C_2\times M_2 \\ 34.1\%\times M_1=13.0\%\times 445 g \\ {M}_1=\frac{13.0 \%\times 445 g}{34.1\%} \\ =169.6 g \end{gathered}$$

 

Hence, consider 169.6 g of 34.1 % (mass) solution and it should be diluted to 445 g with water to get 13.0%(mass) solution.