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The NaNO(3) weighed out to make 50 mL of...

The `NaNO_(3)` weighed out to make 50 mL of an aqueous solution containing 70.0 mg `Na^(+)` per mL is_________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in g `mol^(-1)-Na:23,N:14,O:16` ]

Text Solution

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The correct Answer is:
13
`Na^(+)` present in 50 mL
`= (70 mg)/(1 mL) x 50 mL = 3500 mg = 3.5 g`
Moles of `Na^(+) = 3.5/23`= moles of `NaNO_(3)`
Weight of `NaNO_(3) = 3.5/23 xx 85 = 12.993 g`
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