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Question: How many grams of \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) is needed to make a 0.244 m solut...

How many grams of CH3OH{\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

Explanation

Solution

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 (CH3OH)\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=Moles of soluteMass of solvent\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=Mass of soluteMolar mass of solute\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}} {{{\text{Molar}}\,{\text{mass}}\,{\text{of}}\,{\text{solute}}}}
So, equation (1) becomes,
Molality=Mass of soluteMolar mass of soluteMass of solvent\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 CH3OH{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}
=12+4+16=32 g mol−112 + 4 + 16 = 32\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}
Mass of solvent=400=4001000 kg\dfrac{{400}} {{1000}}\,{\text{kg}}
Molality=0.244 m
Now, we have to substitute the above values in the formula of molality.
0.244 =Mass of solute324001000 0.244\, = \dfrac{{\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}} {{32}}}} {{\dfrac{{400}} {{1000}}\,}}
⇒0.244 mol  kg−1=Mass of solute32 g mol−1×1000400kg\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}}
⇒Mass of solute=0.244×32×4001000=3.12 g\Rightarrow {\text{Mass}}\,{\text{of}}\,{\text{solute}} = \dfrac{{0.244 \times 32 \times 400}} {{1000}} = 3.12\,{\text{g}}
Therefore, the mass of CH3OH{\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.