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
271.2k+ views
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.
Recently Updated Pages
JoSAA Counselling 2026: JoSAA 2026 Mock Seat Allotment LIVE: Round 2 Result Released, Registration, Choice Filling and Ranks

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Understanding the Different Types of Solutions in Chemistry

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Class 12 Chemistry Question Paper 2026 PDF Download (All Sets) with Answer Key

NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2025-26

