
A solution containing 12.5g of non-electrolyte substance in 185g of water shows a boiling point elevation of 0.80 K. Calculate the molar mass of the substance. (${{\text{K}}_{\text{b}}}$=0.52K \[{\text{kgmo}}{{\text{l}}^{{\text{ - 1}}}}\])
A. 53.06 \[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\]
B. 25.3 \[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\]
C. 16.08 \[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\]
D. 43.92 \[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\]
Answer
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Hint: The elevation in boiling point is directly related to the amount of solute added. It is denoted as ${{\Delta }}{{\text{T}}_{\text{b}}}$. It is expressed in Kelvin (K). The molar mass of any substance is \[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\].
Complete step by step answer: The elevation in boiling point, ${{\Delta }}{{\text{T}}_{\text{b}}}$ has the formula which is given below:
${{\Delta }}{{\text{T}}_{\text{b}}}{\text{ = }}{{\text{K}}_{\text{b}}}{\text{ x m}}$
where ${{\text{K}}_{\text{b}}}$is the molal elevation constant, also known as ebullioscopic constant.
m is the molality.
The above equation can be further be written as,
${{\Delta }}{{\text{T}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{1000 x }}{{\text{K}}_{\text{b}}}{\text{ x w}}}}{{{\text{M x W}}}}$
where ‘w’ is the weight of the solute, ‘M’ is the molar mass of the solute and ‘W’ is the weight of the solvent in grams.
It is given that ${{\Delta }}{{\text{T}}_{\text{b}}}$=0.80 K,
W = 185 g
(since 1 kg = 1000 g)
${{\text{K}}_{\text{b}}}$= 0.52K \[{\text{kgmo}}{{\text{l}}^{{\text{ - 1}}}}\]
W = 12.5 g
We need to find the Molar mass of the solute(M).
Rearranging the equation to get molar mass,
${\text{M = }}\dfrac{{{\text{1000 x }}{{\text{K}}_{\text{b}}}{\text{ x w}}}}{{{{ \Delta }}{{\text{T}}_{\text{b}}}{\text{ x W}}}}$
Thus, substituting the given values in the formula we get,
\[
{\text{M = }}\dfrac{{{\text{1000 x 0}}{\text{.52 K kg mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ x 12}}{\text{.5g}}}}{{{\text{ 0}}{\text{.80 x 185 g}}}} \\
\Rightarrow {\text{M = 43}}{\text{.92 g mo}}{{\text{l}}^{{\text{ - 1}}}} \\
\]
The molar mass of the substance is 43.92\[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\].
So, the correct option is D.
Note: The boiling point of the solvent above a solution will be greater than the boiling point of the pure solvent whether the solution contains a non-volatile solute or a volatile solute. in simple words, Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
Complete step by step answer: The elevation in boiling point, ${{\Delta }}{{\text{T}}_{\text{b}}}$ has the formula which is given below:
${{\Delta }}{{\text{T}}_{\text{b}}}{\text{ = }}{{\text{K}}_{\text{b}}}{\text{ x m}}$
where ${{\text{K}}_{\text{b}}}$is the molal elevation constant, also known as ebullioscopic constant.
m is the molality.
The above equation can be further be written as,
${{\Delta }}{{\text{T}}_{\text{b}}}{\text{ = }}\dfrac{{{\text{1000 x }}{{\text{K}}_{\text{b}}}{\text{ x w}}}}{{{\text{M x W}}}}$
where ‘w’ is the weight of the solute, ‘M’ is the molar mass of the solute and ‘W’ is the weight of the solvent in grams.
It is given that ${{\Delta }}{{\text{T}}_{\text{b}}}$=0.80 K,
W = 185 g
(since 1 kg = 1000 g)
${{\text{K}}_{\text{b}}}$= 0.52K \[{\text{kgmo}}{{\text{l}}^{{\text{ - 1}}}}\]
W = 12.5 g
We need to find the Molar mass of the solute(M).
Rearranging the equation to get molar mass,
${\text{M = }}\dfrac{{{\text{1000 x }}{{\text{K}}_{\text{b}}}{\text{ x w}}}}{{{{ \Delta }}{{\text{T}}_{\text{b}}}{\text{ x W}}}}$
Thus, substituting the given values in the formula we get,
\[
{\text{M = }}\dfrac{{{\text{1000 x 0}}{\text{.52 K kg mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ x 12}}{\text{.5g}}}}{{{\text{ 0}}{\text{.80 x 185 g}}}} \\
\Rightarrow {\text{M = 43}}{\text{.92 g mo}}{{\text{l}}^{{\text{ - 1}}}} \\
\]
The molar mass of the substance is 43.92\[{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\].
So, the correct option is D.
Note: The boiling point of the solvent above a solution will be greater than the boiling point of the pure solvent whether the solution contains a non-volatile solute or a volatile solute. in simple words, Boiling-point elevation describes the phenomenon that the boiling point of a liquid (a solvent) will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
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