Suppose two elements `X` and `Y` combine to form two compounds `XY_(2)` and `X_(2)Y_(3)` when `0.05` mole of `XY_(2)` weight `5g` while `3.011xx10^(23)` molecules of `X_(2)Y_(3)` weight `85g`. The atomic masses of `X` and `Y` are respectively.

Options

A. 23,30
B. 30,40
C. 40,30
D. 80,60

Answer

To find the atomic masses of elements `X` and `Y`, we can follow these steps:

### Step 1: Determine the molar mass of the first compound (XY₂)
Given:
- Mass of XY₂ = 5 g
- Moles of XY₂ = 0.05 moles

Using the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
We can rearrange this to find the molar mass:
\[ \text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} \]
Substituting the values:
\[ \text{Molar Mass of XY₂} = \frac{5 \, \text{g}}{0.05 \, \text{moles}} = 100 \, \text{g/mol} \]

### Step 2: Set up the equation for the molar mass of XY₂
The molar mass of XY₂ can be expressed as:
\[ \text{Molar Mass of XY₂} = M(X) + 2M(Y) \]
Where \( M(X) \) is the atomic mass of X and \( M(Y) \) is the atomic mass of Y.
Thus, we have:
\[ M(X) + 2M(Y) = 100 \]
This is our **Equation 1**.

### Step 3: Determine the molar mass of the second compound (X₂Y₃)
Given:
- Number of molecules of X₂Y₃ = \( 3.011 \times 10^{23} \)
- Mass of X₂Y₃ = 85 g

First, calculate the number of moles of X₂Y₃:
Using Avogadro's number \( N_A = 6.022 \times 10^{23} \):
\[ \text{Moles} = \frac{\text{Number of molecules}}{N_A} = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \, \text{moles} \]

Now, using the mass to find the molar mass:
\[ \text{Molar Mass of X₂Y₃} = \frac{85 \, \text{g}}{0.5 \, \text{moles}} = 170 \, \text{g/mol} \]

### Step 4: Set up the equation for the molar mass of X₂Y₃
The molar mass of X₂Y₃ can be expressed as:
\[ \text{Molar Mass of X₂Y₃} = 2M(X) + 3M(Y) \]
Thus, we have:
\[ 2M(X) + 3M(Y) = 170 \]
This is our **Equation 2**.

### Step 5: Solve the system of equations
We now have two equations:
1. \( M(X) + 2M(Y) = 100 \) (Equation 1)
2. \( 2M(X) + 3M(Y) = 170 \) (Equation 2)

From Equation 1, we can express \( M(X) \) in terms of \( M(Y) \):
\[ M(X) = 100 - 2M(Y) \]

Substituting \( M(X) \) into Equation 2:
\[ 2(100 - 2M(Y)) + 3M(Y) = 170 \]
Expanding this gives:
\[ 200 - 4M(Y) + 3M(Y) = 170 \]
Combining like terms:
\[ 200 - M(Y) = 170 \]
Thus:
\[ M(Y) = 200 - 170 = 30 \]

### Step 6: Find M(X)
Now substitute \( M(Y) = 30 \) back into Equation 1:
\[ M(X) + 2(30) = 100 \]
\[ M(X) + 60 = 100 \]
Thus:
\[ M(X) = 100 - 60 = 40 \]

### Final Result
The atomic masses are:
- \( M(X) = 40 \)
- \( M(Y) = 30 \)