Salts of metals X, Y and Z are electrolysed under identical condition using same quantity of electricity . If was observed that 4.2 g of X, 5.4 g of Y and 19.2 g of Z were deposited at respetive cathode . If the atomic weights of X , Y, Z are 7, 27 and 64 respectively , then their ratio of valencies is

Options

A. `1 : 2 : 3`
B. `1 : 3 : 2`
C. `2 : 3 : 1`
D. `3 : 2 : 2`

Answer

To solve the problem, we will use Faraday's second law of electrolysis, which states that the mass of a substance deposited at an electrode during electrolysis is directly proportional to the equivalent mass of that substance. The equivalent mass can be calculated as the molar mass divided by the valency of the metal.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass deposited of metal X (mX) = 4.2 g
- Mass deposited of metal Y (mY) = 5.4 g
- Mass deposited of metal Z (mZ) = 19.2 g
- Atomic weight of X (MW_X) = 7 g/mol
- Atomic weight of Y (MW_Y) = 27 g/mol
- Atomic weight of Z (MW_Z) = 64 g/mol

2. **Set Up the Ratios Using Faraday's Law:**
According to Faraday's law, we can set up the following ratios:
\[ \frac{m_X}{m_Y} = \frac{\frac{MW_X}{n_X}}{\frac{MW_Y}{n_Y}} \implies \frac{m_X}{m_Y} = \frac{MW_X \cdot n_Y}{MW_Y \cdot n_X} \]
Rearranging gives:
\[ \frac{n_X}{n_Y} = \frac{MW_X \cdot m_Y}{MW_Y \cdot m_X} \]

3. **Substituting the Values for X and Y:**
\[ \frac{n_X}{n_Y} = \frac{7 \cdot 5.4}{27 \cdot 4.2} \]
Calculating the right-hand side:
\[ = \frac{37.8}{113.4} = \frac{1}{3} \]
Thus, we have:
\[ n_X : n_Y = 1 : 3 \]

4. **Set Up the Ratios for Y and Z:**
Similarly, we can set up another ratio for metals Y and Z:
\[ \frac{m_Y}{m_Z} = \frac{\frac{MW_Y}{n_Y}}{\frac{MW_Z}{n_Z}} \implies \frac{m_Y}{m_Z} = \frac{MW_Y \cdot n_Z}{MW_Z \cdot n_Y} \]
Rearranging gives:
\[ \frac{n_Y}{n_Z} = \frac{MW_Y \cdot m_Z}{MW_Z \cdot m_Y} \]

5. **Substituting the Values for Y and Z:**
\[ \frac{n_Y}{n_Z} = \frac{27 \cdot 19.2}{64 \cdot 5.4} \]
Calculating the right-hand side:
\[ = \frac{518.4}{345.6} = \frac{3}{2} \]
Thus, we have:
\[ n_Y : n_Z = 3 : 2 \]

6. **Combine the Ratios:**
Now we have:
- \( n_X : n_Y = 1 : 3 \)
- \( n_Y : n_Z = 3 : 2 \)

To find the combined ratio \( n_X : n_Y : n_Z \):
- From \( n_X : n_Y = 1 : 3 \), we can express \( n_Y \) in terms of \( n_X \): \( n_Y = 3n_X \).
- From \( n_Y : n_Z = 3 : 2 \), we can express \( n_Z \) in terms of \( n_Y \): \( n_Z = \frac{2}{3}n_Y = \frac{2}{3} \cdot 3n_X = 2n_X \).

Therefore, we can express all in terms of \( n_X \):
\[ n_X : n_Y : n_Z = n_X : 3n_X : 2n_X = 1 : 3 : 2 \]

### Final Ratio of Valencies:
The ratio of the valencies of metals X, Y, and Z is:
\[ n_X : n_Y : n_Z = 1 : 3 : 2 \]