JEE Main 2021 · 26 February · evening · Numerical answer · +4 / −0
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 and solution
Answer 13
Solution
Chemistry ma’am · 3 steps
Find the total sodium first, then scale it up to the full NaNO3 that carries it.
- Total sodium needed. Each mL holds 70.0 mg of Na⁺, and there are 50 mL. So the solution needs 50 × 70.0 = 3500 mg = 3.5 g of Na⁺.
- Molar mass. NaNO₃ has a molar mass of 23 + 14 + 3 × 16 = 85 g mol⁻¹. One mole of it gives exactly one mole of Na⁺, so 23 g of sodium comes with every 85 g of salt.
m_(NaNO₃) = 3.5 g × 85/23 = 12.93 g
- So. About 12.93 g of NaNO₃ must be weighed out, which rounds to 13.
The common slip: 4. Stopping at the mass of Na⁺ gives 3.5 g, which rounds to 4. That is only the sodium, not the whole NaNO₃.
The idea to keep. When a concentration is given for one ion, get the total mass of that ion first. Then use the formula ratio, here 85/23, to convert it into the mass of the whole salt.
Answer from the NTA final answer key. Solution by PrepWiser.