Question

The density of $2.45M$ aqueous methanol is $0.976{g/ml}$. What is the Molarity of the solution given that the molecular weight of methanol is $32$ grams?
A) $27.3M$
B) $0.273M$
C) $7.23M$
D) $2.73M$

Answer 1

The molecular formula of methanol is $C{H_3}OH$. Given to us, $2.45$ moles of methanol is present in the solution so calculate the weight of methanol present in $2.45$ moles. Then, from the density given, find the weight of the solution.

Complete step-by-step solution: It is given that the density of methanol is $0.976{g/ml}$.
From this information, we can calculate the weight of the solution for one liter as follows.
One liter consists of $1000$ grams. Hence one liter of solution weights $0.976 \times 1000 = 976$ grams.
Now, it is also given to us that one liter contains $2.45$ moles of methanol. We shall now calculate the weight of methanol present in those $2.45$ moles.
Weight of methanol present in $2.45$ moles is $2.45 \times 32 = 78.4$ grams.
This means that $78.4$ grams of methanol is present in $967$ grams of the solution. So the weight of water present would be $976 - 78.4 = 897.6$ grams. This value when converted into kilograms would be $0.8976$ kg.
Now, we write the formula for molarity of the solution containing $2.45$ moles of methanol and $0.8976$ kg of water as follows.
The molarity is the ratio between number of moles of methanol present in the solution to the weight of water in kgs i.e. $M = \dfrac{{Mole{s_{\left( {methanol} \right)}}}}{{Mas{s_{\left( {water} \right)}}}}$
By substituting the above calculated values, we get $M = \dfrac{{2.45}}{{0.8976}} = 2.73$

Hence, the molarity of the solution is $2.73M$ i.e. option (D).

Note: The ratio of the weight of a compound to its gram molecular weight gives the total number of moles of that compound used. So the weight present in a particular number of moles of a compound would be the product of the molecular weight of that compound and the number of moles.