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]
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]
Asked by Shiksha User
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1 Answer
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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+)
mg Na+ will be present in
= 12934.78 mg
= 12.93478 gm
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