To determine how much NaNO₃ must be weighed out to make a 50 mL aqueous solution containing 70 mg of Na⁺ per mL, we can follow these steps:
### Step 1: Calculate the total mass of Na⁺ in the solution
The concentration of Na⁺ is given as 70 mg/mL. For a 50 mL solution, the total mass of Na⁺ can be calculated as follows:
\[
\text{Total mass of Na}^+ = \text{Concentration} \times \text{Volume} = 70 \, \text{mg/mL} \times 50 \, \text{mL} = 3500 \, \text{mg}
\]
### Step 2: Convert the mass of Na⁺ from milligrams to grams
Since 1 mg = 0.001 g, we convert 3500 mg to grams:
\[
\text{Total mass of Na}^+ = 3500 \, \text{mg} \times 0.001 \, \text{g/mg} = 3.5 \, \text{g}
\]
### Step 3: Determine the molar mass of NaNO₃
To find out how much NaNO₃ is needed, we need to know the molar mass of NaNO₃. The molar mass can be calculated as follows:
- Atomic mass of Na = 23 g/mol
- Atomic mass of N = 14 g/mol
- Atomic mass of O = 16 g/mol (and there are 3 oxygen atoms)
\[
\text{Molar mass of NaNO}_3 = 23 + 14 + (3 \times 16) = 23 + 14 + 48 = 85 \, \text{g/mol}
\]
### Step 4: Calculate the percentage composition of Na in NaNO₃
To find out how much NaNO₃ is required to obtain 3.5 g of Na⁺, we need to calculate the percentage of Na in NaNO₃:
\[
\text{Percent composition of Na} = \left( \frac{\text{Mass of Na}}{\text{Molar mass of NaNO}_3} \right) \times 100 = \left( \frac{23}{85} \right) \times 100 \approx 27.06\%
\]
### Step 5: Use the percentage composition to find the mass of NaNO₃ needed
From the percentage composition, we know that 27 g of Na⁺ is present in 100 g of NaNO₃. We can set up a proportion to find out how much NaNO₃ is needed for 3.5 g of Na⁺:
\[
\frac{27 \, \text{g Na}^+}{100 \, \text{g NaNO}_3} = \frac{3.5 \, \text{g Na}^+}{x \, \text{g NaNO}_3}
\]
Cross-multiplying gives:
\[
27x = 350 \implies x = \frac{350}{27} \approx 12.96 \, \text{g}
\]
### Step 6: Round the answer
Rounding 12.96 g gives us approximately 13 g of NaNO₃.
### Final Answer:
To prepare 50 mL of an aqueous solution containing 70 mg of Na⁺ per mL, you need to weigh out approximately **13 g of NaNO₃**.
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