
How many grams of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ is needed to make a 0.244 m solution in 400 g of water?
A.3.1240 g
B.0.313 g
C.1639 g
D.97.6 g
E.32 g
Answer
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Hint: We know that molality is a way of expressing concentration of a solution. Molality is defined as the moles of solute present in 1 kg of solvent. Here, we have to calculate the mass of solute $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}} \right)$ needed to make 0.244 m solution in 400 g of water.
Complete step by step answer:
The molality (m) of the solution is given as 0.244 m. The mass of solvent is 400 g.
We know the formula of molality, that is,
Molality=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}}}$ …… (1)
We know that the moles of solute is,
Moles of solute=$\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{solute}}}}$
So, equation (1) becomes,
Molality=$\dfrac{{\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{solute}}}}}}
{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}}}$
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$
=$12 + 4 + 16 = 32\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Mass of solvent=400=$\dfrac{{400}}
{{1000}}\,{\text{kg}}$
Molality=0.244 m
Now, we have to substitute the above values in the formula of molality.
$0.244\, = \dfrac{{\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{32}}}}
{{\dfrac{{400}}
{{1000}}\,}}$
$ \Rightarrow 0.244\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}} = \dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{32\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}} \times \dfrac{{1000}}
{{400}}{\text{kg}}$
$ \Rightarrow {\text{Mass}}\,{\text{of}}\,{\text{solute}} = \dfrac{{0.244 \times 32 \times 400}}
{{1000}} = 3.12\,{\text{g}}$
Therefore, the mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ required to is to make a 0.244 m solution in 400 g of water is 3.12 g.
So, the correct answer is Option A.
Additional Information:
It is to be noted that molality and molarity are not the same. Molarity is also a way of expressing concentration of a solution. It is defined as the number of moles of solute present in 1 litre of solution. But molality is measured considering the mass of solvent. Molarity changes with the change in temperature but molality value does not change with the temperature.
Note: Always remember that molality is regarded better than molarity for expressing the concentration of a solution because its value does not change. But there is one disadvantage also since molality can be determined only if the density of the solution is known.
Complete step by step answer:
The molality (m) of the solution is given as 0.244 m. The mass of solvent is 400 g.
We know the formula of molality, that is,
Molality=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}}}$ …… (1)
We know that the moles of solute is,
Moles of solute=$\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{solute}}}}$
So, equation (1) becomes,
Molality=$\dfrac{{\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{solute}}}}}}
{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}}}$
Molar mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$
=$12 + 4 + 16 = 32\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Mass of solvent=400=$\dfrac{{400}}
{{1000}}\,{\text{kg}}$
Molality=0.244 m
Now, we have to substitute the above values in the formula of molality.
$0.244\, = \dfrac{{\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{32}}}}
{{\dfrac{{400}}
{{1000}}\,}}$
$ \Rightarrow 0.244\,{\text{mol}}\,\,{\text{k}}{{\text{g}}^{ - 1}} = \dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}}
{{32\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}} \times \dfrac{{1000}}
{{400}}{\text{kg}}$
$ \Rightarrow {\text{Mass}}\,{\text{of}}\,{\text{solute}} = \dfrac{{0.244 \times 32 \times 400}}
{{1000}} = 3.12\,{\text{g}}$
Therefore, the mass of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ required to is to make a 0.244 m solution in 400 g of water is 3.12 g.
So, the correct answer is Option A.
Additional Information:
It is to be noted that molality and molarity are not the same. Molarity is also a way of expressing concentration of a solution. It is defined as the number of moles of solute present in 1 litre of solution. But molality is measured considering the mass of solvent. Molarity changes with the change in temperature but molality value does not change with the temperature.
Note: Always remember that molality is regarded better than molarity for expressing the concentration of a solution because its value does not change. But there is one disadvantage also since molality can be determined only if the density of the solution is known.
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