
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 $32\text{ g}\text{.mol}{{\text{ }}^{-1}}\text{ }$.
Answer
473.4k+ views
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 $\text{ d = }\dfrac{\text{m}}{\text{V}}\text{ }$
$\text{ }\dfrac{\text{1}}{\text{m}}\text{ = }\dfrac{\text{d}}{\text{M}}\text{ }-\text{ }\dfrac{{{\text{M}}_{\text{B}}}}{\text{1000}}\text{ }$
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 $\text{ d = }\dfrac{\text{m}}{\text{V}}\text{ }$
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
