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` ]

Answer

To find the mass of NaNO₃ needed to prepare a 50 mL solution containing 70.0 mg of Na⁺ per mL, we can follow these steps:

### Step 1: Calculate the total amount of Na⁺ in 50 mL
Given that the concentration of Na⁺ is 70.0 mg/mL, we can calculate the total amount of Na⁺ in 50 mL:

\[ \text{Total Na}^+ = \text{Concentration} \times \text{Volume} = 70.0 \, \text{mg/mL} \times 50 \, \text{mL} = 3500 \, \text{mg} \]

### Step 2: Convert the mass of Na⁺ from mg to g
To convert milligrams to grams, we use the conversion factor \(1 \, \text{g} = 1000 \, \text{mg}\):

\[ \text{Total Na}^+ = \frac{3500 \, \text{mg}}{1000} = 3.5 \, \text{g} \]

### Step 3: Calculate the number of moles of Na⁺
Using the atomic weight of sodium (Na), which is 23 g/mol, we can calculate the number of moles of Na⁺:

\[ \text{Number of moles of Na}^+ = \frac{\text{mass}}{\text{molar mass}} = \frac{3.5 \, \text{g}}{23 \, \text{g/mol}} \approx 0.15217 \, \text{mol} \]

### Step 4: Determine the number of moles of NaNO₃
Since each mole of NaNO₃ produces one mole of Na⁺, the number of moles of NaNO₃ will be the same as the number of moles of Na⁺:

\[ \text{Number of moles of NaNO}_3 = 0.15217 \, \text{mol} \]

### Step 5: Calculate the mass of NaNO₃
Now, we need to find the mass of NaNO₃ using its molar mass. The molar mass of NaNO₃ is calculated as follows:

\[ \text{Molar mass of NaNO}_3 = \text{Molar mass of Na} + \text{Molar mass of N} + 3 \times \text{Molar mass of O} = 23 + 14 + 3 \times 16 = 85 \, \text{g/mol} \]

Now we can calculate the mass of NaNO₃:

\[ \text{Mass of NaNO}_3 = \text{Number of moles} \times \text{Molar mass} = 0.15217 \, \text{mol} \times 85 \, \text{g/mol} \approx 12.93 \, \text{g} \]

### Step 6: Round to the nearest integer
Finally, rounding 12.93 g to the nearest integer gives us:

\[ \text{Mass of NaNO}_3 \approx 13 \, \text{g} \]

Thus, the mass of NaNO₃ needed to prepare the solution is **13 g**.

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