The NaNO3 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]
Answer (integer)
13
Solution
Na<sup>+</sup> = 70 mg/mL<br><br>W<sub>Na+</sub> in 50 mL solution<br><br> = 70 $\times$ 50 mg<br><br>= 3500 mg<br><br>= 3.5 gm<br><br>Moles of Na<sup>+</sup> in 50 mL solution = ${{3.5} \over {23}}$<br><br>Moles of NaNO<sub>3</sub> = moles of Na<sup>+</sup><br><br>= ${{3.5} \over {23}}$ mol<br><br>Mass of NaNO<sub>3</sub> = ${{3.5} \over {23}} \times 85 = 12.934$<br><br>$\simeq$ 13 gm
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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