The molar volume of KCl and NaCl are `37.46` mL and `27.94`mL respectively . The ratio of the unit cube edges of the two crystals is

Options

A. `1.296`
B. `1.116`
C. `1.341`
D. `0.950`

Answer

To find the ratio of the unit cube edges of KCl and NaCl, we can follow these steps:

### Step 1: Understand the relationship between volume and edge length
The volume \( V \) of a cubic unit cell is related to the edge length \( a \) by the formula:
\[ V = a^3 \]
Where \( V \) is the volume and \( a \) is the edge length of the cube.

### Step 2: Set up the relationship for KCl and NaCl
Let:
- \( V_1 \) = volume of KCl = 37.46 mL
- \( V_2 \) = volume of NaCl = 27.94 mL
- \( a_1 \) = edge length of KCl
- \( a_2 \) = edge length of NaCl

From the relationship of volume and edge length, we can write:
\[ \frac{V_1}{V_2} = \frac{a_1^3}{a_2^3} \]

### Step 3: Substitute the known volumes into the equation
Substituting the values of \( V_1 \) and \( V_2 \):
\[ \frac{37.46}{27.94} = \frac{a_1^3}{a_2^3} \]

### Step 4: Take the cube root of both sides
To find the ratio of the edge lengths, we take the cube root of both sides:
\[ \frac{a_1}{a_2} = \sqrt[3]{\frac{37.46}{27.94}} \]

### Step 5: Calculate the cube root
Now we calculate the ratio:
\[ \frac{a_1}{a_2} = \sqrt[3]{\frac{37.46}{27.94}} \approx \sqrt[3]{1.34} \approx 1.116 \]

### Step 6: State the final answer
Thus, the ratio of the unit cube edges of KCl to NaCl is:
\[ \frac{a_1}{a_2} \approx 1.116 \]