The molarity of the solution containing 2.8%(m/v) solution of KOH is : (Given atomic mass of K=39) is :

0.1 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.2 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.5 M$
Weight of KOH = $2.8$ gram

Volume of solution = $100$ ml

$M = \frac{2.8 \times 1000}{56 \times 100} = \frac{28}{56} = 0.5 M$

Suggest Corrections

Similar questions

2.8 %

K O H

K = 39 u

The molarity of the solution containing 2.8% mass-volume solution of KOH. (Given atomic mass of K = 39) is

2.8 %

2.8 %

Join BYJU'S Learning Program