A solution containing `12.0%` sodium hydroxide by mass has a density of `1.131 g//mL`. What volume of this solution contains `5.00 mol` of NaOH :

Options

A. `0.0240 L`
B. `1.67 L`
C. `1.47 L`
D. `1.00 L`

Explanation

To solve the problem step by step, we need to find the volume of a sodium hydroxide solution that contains 5.00 moles of NaOH. Here’s how we can do it:

### Step 1: Calculate the mass of NaOH in grams.
To find the mass of NaOH, we use the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
The molar mass of NaOH (sodium hydroxide) is approximately 40 g/mol (Na = 23, O = 16, H = 1). Therefore:
\[ \text{Mass of NaOH} = 5.00 \, \text{mol} \times 40 \, \text{g/mol} = 200 \, \text{g} \]

### Step 2: Use the percentage by mass to find the mass of the solution.
The solution is 12.0% NaOH by mass. This means that in 100 g of the solution, there are 12 g of NaOH. We can set up a proportion to find the total mass of the solution that contains 200 g of NaOH:
\[ \frac{12 \, \text{g NaOH}}{100 \, \text{g solution}} = \frac{200 \, \text{g NaOH}}{x \, \text{g solution}} \]
Cross-multiplying gives:
\[ 12x = 200 \times 100 \]
\[ 12x = 20000 \]
\[ x = \frac{20000}{12} \approx 1666.67 \, \text{g solution} \]

### Step 3: Calculate the volume of the solution using its density.
The density of the solution is given as 1.131 g/mL. We can use the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging this gives us:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values we have:
\[ \text{Volume} = \frac{1666.67 \, \text{g}}{1.131 \, \text{g/mL}} \approx 1471.59 \, \text{mL} \]

### Step 4: Convert the volume from mL to L.
To convert mL to L, we divide by 1000:
\[ \text{Volume in L} = \frac{1471.59 \, \text{mL}}{1000} \approx 1.47159 \, \text{L} \]

### Final Answer:
The volume of the solution that contains 5.00 moles of NaOH is approximately **1.47 L**.

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