Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
63 views
in Chemistry by (92.7k points)
closed by
`45 g` of ethylene glycol `C_(2)H_(6)O_(2)` is mixed with `600 g` of water. Calculate (a) the freezing point depression and (b) the freezing point of solution.
Given`K_(f)=1.86 K kg mol^(-1)`.

1 Answer

0 votes
by (91.9k points)
selected by
 
Best answer
Depression in freezing point is realed to the molality, therefore the molality of the solution with respect to ethylene glycol,
`DeltaT_(f)=K_(f)m`
Mole of ethylene glycol = `45 gxx(1 mol)/(62g) =0.73 mol`
Mass of water in kg =`600 xx (1 kg)/(1000 g)=0.60 kg`
Hence, molality of ethylene glycol
`=(0.73 mol)/(0.60 kg)=1.20 mol kg`
Therefore,freezing point depression
`(DeltaT_(f)=1.86 K kg mol^(-1))xx(1.2 mol kg^(-1) )=2.2 K`
Freezing point of the aqueous solution
`T_(f) =T_(f)^(@) - Delta_(f)T`
`273.15 K -2.2 K =270.95 K`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...