Q13.64 P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molality of the following: 

(a) A solution containing 85.4 g of glycine ( ${\mathbf{NH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{COOH}$) dissolved in 1.270 kg of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. 

(b) A solution containing 8.59 g of glycerol (${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}$ ) in 77.0 g of ethanol  $\left({\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\right)$.

Step-by-Step Solution

Answer

1. The molality of the solution containing 85.4g of glycine, ${\mathbf{NH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{COOH}$  in 1.270kg of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ solvent is 0.897m.
2. The molality of the solution containing 85.4g of glycerol, ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}$  in 0.077kg of ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}$  solvent is 1.21m.

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.

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

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

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

Step 2: Expression to calculate the Concentration

1. Mass of  ${\mathbf{NH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{COOH}\mathbf{=}\mathbf{85}\mathbf{.}\mathbf{4}\mathbf{g}$

Molar Mass of  ${\mathbf{NH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{COOH}\mathbf{=}\mathbf{75}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{=}\frac{\mathbf{85}\mathbf{.}\mathbf{4}\mathbf{g}}{\mathbf{75}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{mol} \end{gathered}$$

Volume of Solvent ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{270}\mathbf{kg}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{solute}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{14}\mathbf{mol}}{\mathbf{1}\mathbf{.}\mathbf{270}\mathbf{kg}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{897}\mathbf{m} \end{gathered}$$

b) Mass of  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{59}\mathbf{g}$

Molar Mass of  ${\mathbf{C}}_{\mathbf{3}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{3}}\mathbf{=}\mathbf{92}\mathbf{g}\mathbf{/}\mathbf{mole}$

$$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Molar} \mathbf{Mass}} \\ \mathbf{=}\frac{\mathbf{8}\mathbf{.}\mathbf{59}\mathbf{g}}{\mathbf{92}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0934} \mathbf{mol} \end{gathered}$$

Volume of Solvent  ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}\mathbf{OH}\mathbf{=}\mathbf{77}\mathbf{g}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{077}\mathbf{kg}$

$$\begin{gathered} \mathbf{Molality}\mathbf{=}\frac{\mathbf{Number} \mathbf{of} \mathbf{Moles} \mathbf{solute}}{\mathbf{Volume} \mathbf{of} \mathbf{the} \mathbf{Solution} \mathbf{(} \mathbf{in} \mathbf{Litre} \mathbf{)}} \\ \mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0934} \mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{077}\mathbf{kg}} \\ \mathbf{=}\mathbf{1}\mathbf{.}\mathbf{21}\mathbf{m} \end{gathered}$$