Answer
414.9k+ views
Hint:
We can use the relationship between the molality of the solution and elevation of boiling point along with that of amount and molar mass.
Complete step by step solution
Among other colligative properties, we have elevation of boiling point of a solvent that also results from the lowering in vapor pressure of the solvent upon addition of a non-volatile solute.
The elevation of boiling point can be related to the molality of the solution by the following expression:
$\Delta {T_b} = {K_b}m$
Here, $\Delta {T_b}$ is the elevation of boiling point of the solvent, ${K_b}$ is the molal
elevation constant for the given solvent and is the molality of the solution.
We can also write the molality of a solution by using number of moles of solute $\left( {{n_{solute}}} \right)$ present in the given mass of solvent $\left( {{m_{solvent}}} \right)$ as follows:
$m = \dfrac{{{n_{solute}}}}{{{m_{solvent}}\;in\;kg}}$
We know that the mass of $1$ mole of a substance is its molar mass $M$ . So we can write:
\[\dfrac{{1\;{\rm{mol}}}}{M}\]
We can use this conversion factor to calculate the amount of solute in given mass \[\left( {{m_{solute}}} \right)\] by using its molar mass \[\left( {{M_{solute}}} \right)\]as follows:
\[
{n_{solute}} = {m_{solute}} \times \left( {\dfrac{{1\;{\rm{mol}}}}{{{M_{solute}}\;g}}} \right)\\
= \dfrac{{{m_{solute}}}}{{{M_{solute}}}}
\]
Let’s rewrite the previously written expression of molality by using the above expression for amount of solute as follows:
\[m = \dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}\]
We can use this equation to rewrite the expression for elevation of boiling point as follows:
$\Delta {T_b} = {K_b}\left\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\}$
Let’s rearrange this equation for molar mass of solute:
$
{K_b}\left\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\} = \Delta {T_b}\\
\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}} = \dfrac{{\Delta {T_b}}}{{{K_b}}}\\
{M_{solute}} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_{solute}}}}{{{m_{solvent}}\;in\;kg}}
$
Here, we have used subscript $1$ for solvent and \[2\] for solute. So, let’s incorporate these in our derived equation as follows:
${M_2} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_2}}}{{{m_1}}}$
Hence, the correct option is C.
Note:
We have to use the subscripts carefully for solute and solvent in the expressions.
We also have to take care of the units as well.
We can use the relationship between the molality of the solution and elevation of boiling point along with that of amount and molar mass.
Complete step by step solution
Among other colligative properties, we have elevation of boiling point of a solvent that also results from the lowering in vapor pressure of the solvent upon addition of a non-volatile solute.
The elevation of boiling point can be related to the molality of the solution by the following expression:
$\Delta {T_b} = {K_b}m$
Here, $\Delta {T_b}$ is the elevation of boiling point of the solvent, ${K_b}$ is the molal
elevation constant for the given solvent and is the molality of the solution.
We can also write the molality of a solution by using number of moles of solute $\left( {{n_{solute}}} \right)$ present in the given mass of solvent $\left( {{m_{solvent}}} \right)$ as follows:
$m = \dfrac{{{n_{solute}}}}{{{m_{solvent}}\;in\;kg}}$
We know that the mass of $1$ mole of a substance is its molar mass $M$ . So we can write:
\[\dfrac{{1\;{\rm{mol}}}}{M}\]
We can use this conversion factor to calculate the amount of solute in given mass \[\left( {{m_{solute}}} \right)\] by using its molar mass \[\left( {{M_{solute}}} \right)\]as follows:
\[
{n_{solute}} = {m_{solute}} \times \left( {\dfrac{{1\;{\rm{mol}}}}{{{M_{solute}}\;g}}} \right)\\
= \dfrac{{{m_{solute}}}}{{{M_{solute}}}}
\]
Let’s rewrite the previously written expression of molality by using the above expression for amount of solute as follows:
\[m = \dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}\]
We can use this equation to rewrite the expression for elevation of boiling point as follows:
$\Delta {T_b} = {K_b}\left\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\}$
Let’s rearrange this equation for molar mass of solute:
$
{K_b}\left\{ {\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}}} \right\} = \Delta {T_b}\\
\dfrac{{{m_{solute}}}}{{\left( {{M_{solute}}} \right)\left( {{m_{solvent}}\;in\;kg} \right)}} = \dfrac{{\Delta {T_b}}}{{{K_b}}}\\
{M_{solute}} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_{solute}}}}{{{m_{solvent}}\;in\;kg}}
$
Here, we have used subscript $1$ for solvent and \[2\] for solute. So, let’s incorporate these in our derived equation as follows:
${M_2} = \dfrac{{{K_b}}}{{\Delta {T_b}}}\dfrac{{{m_2}}}{{{m_1}}}$
Hence, the correct option is C.
Note:
We have to use the subscripts carefully for solute and solvent in the expressions.
We also have to take care of the units as well.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)