What volume of `NH_(3)` gas at STP would be needed to prepare 100 ml of 2.5 molal (2.5 m) ammonium hydroxide solution?

Options

A. (a)0.056 litre
B. (b)0.56 litre
C. (c )5.6 litres
D. (d)11.2 litres

Explanation

To solve the problem of how much volume of `NH3` gas at STP is needed to prepare 100 ml of a 2.5 molal (2.5 m) ammonium hydroxide solution, we can follow these steps:

### Step 1: Understand the definition of molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, we need to find the number of moles of ammonium hydroxide (NH4OH) that can be prepared from the given volume and molality.

### Step 2: Calculate the mass of the solvent (water)
Given that the volume of the solvent (water) is 100 ml and the density of water is approximately 1 g/ml, we can calculate the mass of water:
\[ \text{Mass of water} = \text{Volume} \times \text{Density} = 100 \, \text{ml} \times 1 \, \text{g/ml} = 100 \, \text{g} \]
Convert this mass into kilograms:
\[ \text{Mass of water in kg} = \frac{100 \, \text{g}}{1000} = 0.1 \, \text{kg} \]

### Step 3: Calculate the number of moles of solute
Using the definition of molality:
\[ \text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
We rearrange this to find the moles of solute:
\[ \text{Moles of solute} = \text{Molality} \times \text{mass of solvent in kg} = 2.5 \, \text{mol/kg} \times 0.1 \, \text{kg} = 0.25 \, \text{mol} \]

### Step 4: Relate moles of NH4OH to moles of NH3
To prepare ammonium hydroxide (NH4OH), we need ammonia (NH3) and water (H2O). The reaction shows that 1 mole of NH3 produces 1 mole of NH4OH. Therefore, to prepare 0.25 moles of NH4OH, we need:
\[ \text{Moles of NH3 required} = 0.25 \, \text{mol} \]

### Step 5: Calculate the volume of NH3 at STP
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Thus, the volume of 0.25 moles of NH3 is:
\[ \text{Volume of NH3} = \text{Moles of NH3} \times \text{Volume per mole at STP} = 0.25 \, \text{mol} \times 22.4 \, \text{L/mol} = 5.6 \, \text{L} \]

### Final Answer
The volume of `NH3` gas needed at STP to prepare 100 ml of a 2.5 molal ammonium hydroxide solution is **5.6 liters**.

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