Calculate the mass of potassium sulfate required to prepare its 33% solution in 100g of water.

Options

A. 49g
B. 14g
C. 26g
D. 28g

Explanation

To calculate the mass of potassium sulfate (K₂SO₄) required to prepare a 33% solution in 100 grams of water, we can follow these steps:

### Step 1: Understand the percentage concentration
The percentage concentration (mass/mass %) is defined as:
\[ \text{mass \%} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \]
In this case, we want a 33% solution.

### Step 2: Define variables
Let:
- \( X \) = mass of potassium sulfate (solute) in grams
- The mass of water (solvent) = 100 g
- The mass of the solution = mass of solute + mass of solvent = \( X + 100 \) g

### Step 3: Set up the equation
Using the definition of mass percentage:
\[ 33 = \left( \frac{X}{X + 100} \right) \times 100 \]

### Step 4: Rearrange the equation
To eliminate the percentage, we can simplify the equation:
\[ \frac{X}{X + 100} = \frac{33}{100} \]

### Step 5: Cross-multiply to solve for \( X \)
Cross-multiplying gives:
\[ 100X = 33(X + 100) \]

### Step 6: Distribute and simplify
Expanding the right side:
\[ 100X = 33X + 3300 \]

### Step 7: Isolate \( X \)
Now, subtract \( 33X \) from both sides:
\[ 100X - 33X = 3300 \]
\[ 67X = 3300 \]

### Step 8: Solve for \( X \)
Now, divide both sides by 67:
\[ X = \frac{3300}{67} \approx 49.25 \text{ g} \]

### Step 9: Conclusion
Thus, the mass of potassium sulfate required to prepare a 33% solution in 100 g of water is approximately **49.25 grams**.

---