An alloy consists of rubidium and one of the other alkali metals. A sample of `4.6g` of the alloy when allowed to react with water, liberates `2.241dm^(3)` of hydrogen at `STP`.
Relative atomic masses: `A_(T)(Li)=7,A_(T)(Na)=23,A_(T)(K)=39,A_(T)(Rb)=85.5,A_(T)(Cs)=1.33`
What composition in `%` by mass has the alloy?

Answer

To solve the problem step by step, we will follow the outlined reasoning and calculations from the video transcript.

### Step 1: Write the Reaction Equation
The general reaction of an alkali metal (M) with water is:
\[ M + H_2O \rightarrow MOH + \frac{1}{2}H_2 \]
This means that 2 moles of metal react with 2 moles of water to produce 2 moles of metal hydroxide and 1 mole of hydrogen gas.

### Step 2: Calculate Moles of Hydrogen Gas
Given that the volume of hydrogen gas liberated is \(2.241 \, \text{dm}^3\) at STP (Standard Temperature and Pressure), we can calculate the number of moles of hydrogen gas using the molar volume of a gas at STP, which is \(22.4 \, \text{dm}^3/\text{mol}\):
\[ \text{Moles of } H_2 = \frac{2.241 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.1 \, \text{mol} \]

### Step 3: Relate Moles of Metal to Moles of Hydrogen
From the balanced equation, we see that 2 moles of metal react to produce 1 mole of hydrogen. Therefore, the moles of metal (M) can be calculated as:
\[ \text{Moles of metal} = 2 \times \text{Moles of } H_2 = 2 \times 0.1 = 0.2 \, \text{mol} \]

### Step 4: Calculate Mean Molar Mass of the Alloy
The mean molar mass of the alloy can be calculated using the total mass of the alloy and the number of moles of metal:
\[ \text{Mean Molar Mass} = \frac{\text{Mass of alloy}}{\text{Moles of metal}} = \frac{4.6 \, \text{g}}{0.2 \, \text{mol}} = 23 \, \text{g/mol} \]

### Step 5: Identify the Constituent Metals
We know one of the metals is rubidium (Rb) with an atomic mass of \(85.5 \, \text{g/mol}\). The mean molar mass of \(23 \, \text{g/mol}\) suggests that the other metal must be lithium (Li), which has an atomic mass of \(7 \, \text{g/mol}\).

### Step 6: Set Up Equations for Moles and Mass
Let \(n_{Rb}\) be the number of moles of rubidium and \(n_{Li}\) be the number of moles of lithium. We have two equations:
1. \(n_{Rb} + n_{Li} = 0.2\)
2. \(85.5 \cdot n_{Rb} + 7 \cdot n_{Li} = 4.6\)

### Step 7: Solve the Equations
From equation 1, we can express \(n_{Li}\) in terms of \(n_{Rb}\):
\[ n_{Li} = 0.2 - n_{Rb} \]
Substituting this into equation 2:
\[ 85.5 \cdot n_{Rb} + 7 \cdot (0.2 - n_{Rb}) = 4.6 \]
Expanding and simplifying:
\[ 85.5 \cdot n_{Rb} + 1.4 - 7 \cdot n_{Rb} = 4.6 \]
\[ 78.5 \cdot n_{Rb} = 4.6 - 1.4 \]
\[ 78.5 \cdot n_{Rb} = 3.2 \]
\[ n_{Rb} = \frac{3.2}{78.5} \approx 0.0408 \, \text{mol} \]

### Step 8: Calculate Moles of Lithium
Using \(n_{Rb}\) to find \(n_{Li}\):
\[ n_{Li} = 0.2 - 0.0408 = 0.1592 \, \text{mol} \]

### Step 9: Calculate Percentage Composition
Now, we can calculate the mass of each metal:
- Mass of rubidium:
\[ \text{Mass}_{Rb} = n_{Rb} \cdot 85.5 = 0.0408 \cdot 85.5 \approx 3.49 \, \text{g} \]
- Mass of lithium:
\[ \text{Mass}_{Li} = n_{Li} \cdot 7 = 0.1592 \cdot 7 \approx 1.11 \, \text{g} \]

### Step 10: Calculate Percentage by Mass
- Percentage of rubidium:
\[ \text{Percentage}_{Rb} = \left(\frac{3.49}{4.6}\right) \times 100 \approx 75.83\% \]
- Percentage of lithium:
\[ \text{Percentage}_{Li} = \left(\frac{1.11}{4.6}\right) \times 100 \approx 24.22\% \]

### Final Result
The composition of the alloy is approximately:
- Rubidium: \(75.83\%\)
- Lithium: \(24.22\%\)