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]

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1 Answer
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6 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        

 

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