LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
37 views
in Chemistry by (90.5k points)
closed by
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K and freezing point of pure water is 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is :
A. 271 K
B. 273.15
C. 269.07 K
D. 277.23 K.

1 Answer

0 votes
by (90.8k points)
selected by
 
Best answer
Correct Answer - c
For both the solution `W_(B)andW_(A)` are the same.
`(DeltaT_(f(g)))/(DeltaT_(f(s)))=(M_(("suorse")))/(M_(("glucose")))`
`DeltaT_(f(g))=((342g mol^(-1)))/((182g mol^(-1)))xx215K=4.04 K`
`"f.p. of glucose solution"=(273.15-4.-4)`
=269.11 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

...