A 7 M solution of potassium hydroxide (KOH) in water contains 40% by weight of KOH. The density of the solution is

Options

A. 1.96
B. 1.28
C. 0.49
D. 0.98

Explanation

To find the density of a 7 M solution of potassium hydroxide (KOH) that contains 40% by weight of KOH, we can follow these steps:

### Step 1: Understand the given data
- Molarity (M) = 7 M
- Percentage by weight of KOH = 40%
- Molar mass of KOH = 56 g/mol

### Step 2: Calculate the mass of KOH in the solution
Since the solution is 40% by weight KOH, this means that in 100 grams of the solution, there are 40 grams of KOH.

### Step 3: Calculate the number of moles of KOH
To find the number of moles of KOH, we use the formula:
\[ \text{Moles of KOH} = \frac{\text{mass of KOH}}{\text{molar mass of KOH}} \]
Substituting the values:
\[ \text{Moles of KOH} = \frac{40 \text{ g}}{56 \text{ g/mol}} = \frac{5}{7} \text{ moles} \]

### Step 4: Calculate the volume of the solution
We know that density (D) is defined as mass per unit volume. Therefore, the volume of the solution can be expressed as:
\[ \text{Volume} = \frac{\text{mass of solution}}{\text{density}} = \frac{100 \text{ g}}{D \text{ g/mL}} \]
To convert this volume from mL to liters, we divide by 1000:
\[ \text{Volume in liters} = \frac{100 \text{ g}}{D \text{ g/mL}} \times \frac{1 \text{ mL}}{1000 \text{ L}} = \frac{100}{D \times 1000} = \frac{1}{10D} \text{ L} \]

### Step 5: Set up the molarity equation
Molarity (M) is defined as moles of solute per liter of solution:
\[ M = \frac{\text{moles of KOH}}{\text{volume of solution in liters}} \]
Substituting the values we have:
\[ 7 = \frac{\frac{5}{7}}{\frac{1}{10D}} \]

### Step 6: Solve for density (D)
Rearranging the equation gives:
\[ 7 = \frac{5 \cdot 10D}{7} \]
Multiplying both sides by 7:
\[ 49 = 50D \]
Now, solving for D:
\[ D = \frac{49}{50} = 0.98 \text{ g/mL} \]

### Final Answer
The density of the solution is **0.98 g/mL**.

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