How much `NaNO_3` must be weighed out to make 50 ml of an aqueous solution containing 70 mg `Na^(+)` per ml?

Options

A. 14 gm
B. 13 gm
C. 18 gm
D. 27 gm

Answer

To solve the problem of how much NaNO₃ must be weighed out to make 50 ml of an aqueous solution containing 70 mg Na⁺ per ml, we will follow these steps:

### Step 1: Calculate the total amount of Na⁺ in 50 ml of solution
Given that the solution contains 70 mg of Na⁺ per ml, we can calculate the total amount of Na⁺ in 50 ml.

\[ \text{Total Na}^+ = 70 \, \text{mg/ml} \times 50 \, \text{ml} = 3500 \, \text{mg} \]

### Step 2: Convert mg of Na⁺ to grams
Since we need to work in grams, we convert milligrams to grams.

\[ 3500 \, \text{mg} = 3500 \times 10^{-3} \, \text{g} = 3.5 \, \text{g} \]

### Step 3: Calculate the molar mass of NaNO₃
Next, we need to find the molar mass of NaNO₃. The molar mass can be calculated using the atomic masses from the periodic table:

- Sodium (Na): 23 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (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
Now, we find the percentage of Na in NaNO₃:

\[ \text{Percentage of Na} = \left( \frac{\text{Molar mass of Na}}{\text{Molar mass of NaNO}_3} \right) \times 100 = \left( \frac{23}{85} \right) \times 100 \approx 27.06\% \]

### Step 5: Use the percentage to find the mass of NaNO₃ required
We know that 27.06 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:

\[ \text{Let } x \text{ be the mass of NaNO}_3. \]

\[ \frac{3.5 \, \text{g Na}}{27.06 \, \text{g Na}} = \frac{x}{100 \, \text{g NaNO}_3} \]

Cross-multiplying gives:

\[ 3.5 \times 100 = 27.06 \times x \]

\[ 350 = 27.06x \]

Solving for \(x\):

\[ x = \frac{350}{27.06} \approx 12.93 \, \text{g} \]

### Step 6: Round to appropriate significant figures
Rounding to two decimal places, we find:

\[ x \approx 12.93 \, \text{g} \approx 13 \, \text{g} \]

### Conclusion
Therefore, the mass of NaNO₃ that must be weighed out is approximately **13 grams**.

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