The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0mg Na+ per mL is…………. g. (Rounded off to the nearest integer)

[Given ; Atomic weight in g mol-1 -Na : 23; N : 14; O : 16]

3 Views|Posted 4 months ago
Asked by Shiksha User
1 Answer
A
4 months ago

Mass of Na+ in 50 ml = 70 * 50 = 3500 mg

23000 mg of Na+ is present in 85000 mg of NaNO3 (1 mole NaNO3 contains 1 mole Na+)

3 5 0 0   mg Na+ will be present in   8 5 0 0 0 2 3 0 0 0 * 3 5 0 0

= 12934.78 mg

= 12.93478 gm

1 3          

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Chemistry Redox Reactions 2025

Chemistry Redox Reactions 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering