CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the volume of 0.015 M HCl solution required to prepare 250 mL of a 5.25×103 M HCl solution.

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

The correct option is B 87.5 mL
M1V1=M2V2
Where,
M1 The molarity of the initial solution = 0.015 M
V1 The volume of the initial solution
M2 The molarity of the resultant solution = (5.25×103) M
V2The volume of the resultant solution = 250 mL
0.015M×V1=5.25×103M×250 mL
V1=5.25×103×2500.015
= 87.5 mL

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Reactions in Solutions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon