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 ]

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answer is 13.

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Detailed Solution

NaNO3→Na++NO3− 1mole 85g⟶23g85mg⟶23mg?⟶70mg=70×8523mgMass of NaNO3 required for 1 mL of solution =70×8523mgFor 50 mL of solution, mass of NaNO3=70×8523×50=12934 mg=12.93 g≈13 g

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