LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
27 views
in Chemistry by (71.2k points)
closed by
A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.

1 Answer

0 votes
by (117k points)
selected by
 
Best answer
`Delta T = (1000 xx K xx w)/(m xx W)`
Assuming `5%` weight per cent, we have
`w = 5g, W = 100-5 = 95g`,
`Delta T = 273.15-271 = 2.15, m = 342` (for sugar)
`therefore 2.15 = (1000 xx K xx 5)/(342 xx 95)` …(1)
For glucose `Delta T = (1000 xx K xx 5)/(180 xx 95)` ...(2)
By eqs. (1) and (2), `Delta T = 4.085`
`therefore` Freezing point `= 273.15-4.09`
`=269.06 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

...