The boiling point of an aqueous solution is found to be 100.280C. Then the freezing point of the same solution is –x0C. What is the value of ‘x’?

Solution:

$\begin{array}{l} △ T_{b} = 0 .28 = K_{b} \times {molarityK}_{b} = 0 .52 for water \Rightarrow \frac{0 .28}{0 .52} = molarity molarity = 0 .5384 \\ △ T_{f} = 1 .86 \times molarity △ T_{f} = 1 .86 \times 0 .5384 △ T_{f} = 1 .00 \\ T_{f} = - 1^{0} C \end{array}$

Related content

World Vegetarian Day

Asia Cup Winners List 1984-2023, Country Wise, Runners Up, Captains, Hosts

NEET 2024 Roadmap: Tips to Know How to Prepare for NEET 2024

TANGEDCO Full Form – Tamil Nadu Generation and Distribution Corporation

Guru Nanak Ji Biography

Best Medical Courses to Pursue Without NEET Exam After 12th in 2023-24

Sardar Vallabhbhai Patel Biography

Airbnb Full Form

COVID-19 Essay in English

DU (Delhi University) 2023-24 Scholarship List and Opportunity for All Students