Competishun
Contact Us
Back

Report Issue

JEE MAIN 2025
28-01-2025 SHIFT-1
Question
The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ $ \times {10^{ - 1}}$ M(Nearest integer)
[Given: Density of aqueous solution of (X) is $1.25\;{\rm{g}}\;{\rm{m}}{{\rm{L}}^{ - 1}}$
Molar mass of the acid is $70\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}$]
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Assuming 100 gm solution contain 70 gm solute. Volume of 100 gm solution will be $\frac{{100}}{{1.25}}{\rm{ml}}$.
${\rm{ Molarity }} = \frac{{70/70}}{{100/1.25}} \times 1000 = 12.5{\rm{ or }}125 \times {10^{ - 1}}$
Question Tags
JEE Main
Chemistry
Easy
Start Preparing for JEE with Competishun
YouTube
Books
App
Tests
Video Solution
BY competishun
Video Solution
Watch Solution