Calculate the molecular weight of a gas `X` which diffuses four times as fast as another gas `Y`, which in turn diffuses twice as fast as another `Z`. Molecular weight of the gas `Z` is `128`.

Answer

To calculate the molecular weight of gas X, we can follow these steps:

### Step 1: Understand the relationship between the rates of diffusion of the gases
We know that:
- Gas X diffuses 4 times as fast as gas Y.
- Gas Y diffuses 2 times as fast as gas Z.

Let's denote:
- Rate of diffusion of gas X as \( R_X \)
- Rate of diffusion of gas Y as \( R_Y \)
- Rate of diffusion of gas Z as \( R_Z \)

From the problem, we can express the relationships as:
\[ R_X = 4 R_Y \]
\[ R_Y = 2 R_Z \]

### Step 2: Use the molecular weight of gas Z
We are given that the molecular weight of gas Z (\( M_Z \)) is 128.

### Step 3: Relate the rates of diffusion to molecular weights using Graham's law
According to Graham's law of diffusion:
\[ \frac{R_Y}{R_X} = \frac{M_X}{M_Y} \]
\[ \frac{R_Y}{R_Z} = \frac{M_Z}{M_Y} \]

### Step 4: Substitute the relationships into the equations
From \( R_Y = 2 R_Z \), we can express it in terms of \( R_Z \):
\[ R_Y = 2 R_Z \]
Thus,
\[ \frac{R_Y}{R_Z} = 2 \]
Using Graham's law:
\[ 2 = \frac{M_Z}{M_Y} \]
Substituting \( M_Z = 128 \):
\[ 2 = \frac{128}{M_Y} \]
From this, we can solve for \( M_Y \):
\[ M_Y = \frac{128}{2} = 64 \]

### Step 5: Now relate \( R_X \) and \( R_Y \) to find \( M_X \)
Using the relationship \( R_X = 4 R_Y \):
\[ \frac{R_X}{R_Y} = 4 \]
Using Graham's law again:
\[ 4 = \frac{M_Y}{M_X} \]
Substituting \( M_Y = 64 \):
\[ 4 = \frac{64}{M_X} \]
From this, we can solve for \( M_X \):
\[ M_X = \frac{64}{4} = 16 \]

### Final Answer
The molecular weight of gas X is \( 16 \).

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