Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

30 mL of CH3OH (d = 0.8 g/cm3) is mixed with 60 mL of C2H5OH (d = 0.92 g/cm3) at 25°C to form a solution of density 0.88 g/cm3. Select the correct option(s) :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

Molarity and molality of resulting solution are 6.33 M and 13.59 m respectively.

b

The mole fraction of solute and molality are 0.38 and 13.59 m respectively.

c

Molarity and percentage change in volume are 13.5 M and zero respectively.

d

Mole fraction of solvent and molality are 0.62 and 13.59 m respectively.

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

30 mL of CH3OH ............CH3OH is solute (less amount)Mass of CH3OH = 30 × 0.8 = 24 gMass of C2H5OH = 60 × 0.92 = 55.2 gMass of solution = 79.2 g Volume of solution =79.20.88=90mL Molarity =nCH0OV(L)=24/3290×1000=8.33M Molality =24/3255.2×1000=13.59m Mole fraction of solute =24322432+55.246=0.38 Mole fraction of solvent =1−0.38=0.62
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring