A bottle of concentrated sulphuric acid (density 1.80 g `cm^(-3)`) is labelled as 86% by weight. What is the molarity of the solution ?

Explanation

To find the molarity of the concentrated sulfuric acid solution, we can follow these steps:

### Step 1: Understand the given data
- Density of the solution = 1.80 g/cm³
- Concentration of sulfuric acid = 86% by weight

### Step 2: Assume a mass for the solution
To simplify calculations, we can assume the total mass of the solution is 100 grams. This makes it easier to calculate the mass of sulfuric acid in the solution.

### Step 3: Calculate the mass of sulfuric acid
Since the solution is 86% sulfuric acid by weight:
- Mass of H₂SO₄ = 86% of 100 g = 86 g

### Step 4: Calculate the number of moles of sulfuric acid
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of H₂SO₄ (sulfuric acid) is approximately 98.079 g/mol. Therefore:
\[ \text{Number of moles of H₂SO₄} = \frac{86 \text{ g}}{98.079 \text{ g/mol}} \approx 0.8768 \text{ moles} \]

### Step 5: Calculate the volume of the solution
Using the density formula:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]
We can rearrange this to find the volume:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \text{ g}}{1.80 \text{ g/cm}^3} \approx 55.56 \text{ cm}^3 \]
Since 1 cm³ = 1 mL, this is equivalent to 55.56 mL. To convert this to liters:
\[ \text{Volume in liters} = \frac{55.56 \text{ mL}}{1000} = 0.05556 \text{ L} \]

### Step 6: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
Substituting the values we calculated:
\[ \text{Molarity} = \frac{0.8768 \text{ moles}}{0.05556 \text{ L}} \approx 15.78 \text{ M} \]

### Final Answer
The molarity of the concentrated sulfuric acid solution is approximately **15.78 M**.

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