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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 $\text{ d = }\dfrac{\text{m}}{\text{V}}\text{ }$

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