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Calculate the number of moles of methyl alcohol in 2 molal solutions in $\text{5 L}$. The density of the solution is $0.981\text{ kg}\text{.d}{{\text{m}}^{-3}}\text{ }$. Given that, the molar mass of methyl alcohol is $32\text{ g}\text{.mol}{{\text{ }}^{-1}}\text{ }$.

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Last updated date: 20th Jun 2024
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Answer
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Hint: The molarity of the solution can be related to the molarity of the solution, the density of the solution, the molar mass of solute as follows,
$\text{ }\dfrac{\text{1}}{\text{m}}\text{ = }\dfrac{\text{d}}{\text{M}}\text{ }-\text{ }\dfrac{{{\text{M}}_{\text{B}}}}{\text{1000}}\text{ }$
Where m is the molality, M is the molarity, d is the density of the solution, and $\text{ }{{\text{M}}_{\text{B}}}\text{ }$ is the molar mass of the solute.

Complete step by step answer:
The molarity is defined as the number of moles dissolved per unit volume in liter.it is used to determine the concentrations of a solution. The molarity is expressed as follows:
$\text{ molarity (M) = }\dfrac{\text{no}\text{. of moles}}{\text{Volume in liter}}\text{ = }\dfrac{\text{n}}{\text{V}}\text{ }$

We have given the following data:
Molality (m) of the solution is 2 molal, $\text{ m = 2 molal }$
The volume of the solution is 5 litre, $\text{ V = 5 L }$
The density of the solution is $\text{ d = }0.981\text{ kg}\text{.d}{{\text{m}}^{-3}}\text{ }$
The molar mass of methyl alcohol is $\text{ M= }32\text{ g}\text{.mol}{{\text{ }}^{-1}}\text{ }$
We have to find the number of moles of methyl alcohol.
The density, molarity (M), molality, and the molar mass of the solute are related as follows,
$\text{ }\dfrac{\text{1}}{\text{m}}\text{ = }\dfrac{\text{d}}{\text{M}}\text{ }-\text{ }\dfrac{{{\text{M}}_{\text{B}}}}{\text{1000}}\text{ }$

Where m is the molality, M is the molarity, d is the density of the solution, and $\text{ }{{\text{M}}_{\text{B}}}\text{ }$ is the molar mass of solute.
Let’s substitute the values to find the molarity of solution. We have,
$\begin{align}
  & \text{ }\dfrac{\text{1}}{2}\text{ = }\dfrac{0.981}{\text{M}}\text{ }-\text{ }\dfrac{32}{\text{1000}}\text{ } \\
 & \Rightarrow \dfrac{0.981}{\text{M}}\text{ = }\dfrac{\text{1}}{2}\text{ + }\dfrac{32}{\text{1000}}\text{ } \\
 & \Rightarrow \dfrac{0.981}{\text{M}}=\text{ }\dfrac{1064}{2000} \\
 & \Rightarrow \text{M = }\dfrac{0.981\text{ }\times \text{ }2000\text{ }}{1064}\text{ = 1}\text{.843 M} \\
\end{align}$
Therefore, the molarity of the solution is equal to the $\text{1}\text{.843 M}$.
Since we know that molarity is related to the number of moles. The number of moles can be calculated as,
$\begin{align}
 & \text{ molarity (M) = }\dfrac{\text{n}}{\text{V}}\text{ } \\
 & \therefore \text{ n = M }\times \text{ V } \\
 & \Rightarrow \text{n=}1.83\text{ }\times \text{ 5L = 9}\text{.15 moles} \\
\end{align}$
Therefore, the number of moles of the methyl alcohol solutions are equal to $\text{9}\text{.15}$.

Note: Note that, molality and molarity are the terms to express the concentrations. For dilute solution, the molarity is equal to molality. Since the density of the solutions are close to $\text{ 1}\text{.0 g/mL}$ as, the volume 1 litre is nearly equal to mass 1 kilogram. For solution other than the water the molality and molarity are different and they can be related by the density relation $\text{ d = }\dfrac{\text{m}}{\text{V}}\text{ }$