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’?

△Tb=0.28=Kb×molarityKb=0.52 for water⇒0.280.52=molarity molarity=0.5384△Tf=1.86×molarity△Tf=1.86×0.5384△Tf=1.00Tf=−10C

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

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’?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

Watch 3-min video & get full concept clarity

hear from our champions

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET