Two sulfuric acid solutions art mixed as follows: $25.0 m t$ . of a $0.50 M$ sulfuric acid solution are added to $0.075 L$ of a $0.25 M$ sulfuric acid solution. What is the molarity of the muhing mixture? (Assume volumes arc additive)

0.501 M

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

0.312 M

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

0.712 M

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

0.120 M

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

Open in App

Solution

The correct option is B $0.312 M$
$M_{1} = 0.50 M$
$V_{1} = \frac{25}{1000} = 0.025 L$
$M_{2} = 0.25 M$
$V_{2} = 0.075 L$
$V = 0.025 + 0.075 = 0.100 L$

Final concentration
$= \frac{M_{1} V_{1} + M_{2} V_{2}}{Total volume}$
$= \frac{(0.5 \times 0.025) + (0.25 \times 0.075)}{0.1}$
$= \frac{0.0125 + 0.01875}{0.1}$
$= 0.3125 M$

Suggest Corrections

Similar questions

N a O H

_{2}

_{4}

d m^{- 3}

Join BYJU'S Learning Program