wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

During the experiment you are given 2.5 m acetone solution in heptane (molar mass = 100 g/mol) having a density of 0.5 g/mL. Find the mole fraction of solute.

A
0.3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
0.85
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.02
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.2
Let us assume the weight of the solvent as 1 kg or 1000 g.
Molality (m)=Moles of solute Weight of solvent (kg)
Moles of acetone (solute) = 2.5 moles
Moles of solvent (heptane)=1000100=10 mol
Mole fraction of solute=Moles of soluteTotal moles=2.5(2.5+10)=2.512.5=0.2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon