**Δng** is the change in the number of moles of gaseous substances during a chemical reaction. To determine the value of Δng for the combustion of 1 mol of liquid ethanol at 298K, we need to consider the reactants and products involved in the reaction.

**Reactants:**
- Liquid ethanol (C2H5OH)

**Products:**
- Carbon dioxide (CO2)
- Water vapor (H2O)

**Step 1: Write the Balanced Chemical Equation**
The balanced chemical equation for the combustion of ethanol is:

C2H5OH + 3O2 → 2CO2 + 3H2O

This equation shows that 1 mol of liquid ethanol reacts with 3 mol of oxygen gas to produce 2 mol of carbon dioxide gas and 3 mol of water vapor.

**Step 2: Calculate the Change in the Number of Moles**
To find the change in the number of moles of gaseous substances, we need to compare the number of moles of gaseous products with the number of moles of gaseous reactants.

**Reactants (gaseous):**
- Oxygen (O2): 3 mol

**Products (gaseous):**
- Carbon dioxide (CO2): 2 mol
- Water vapor (H2O): 3 mol

**Step 3: Calculate the Change in Moles**
The change in the number of moles (Δng) can be calculated by subtracting the total number of moles of gaseous reactants from the total number of moles of gaseous products.

Δng = (2 mol + 3 mol) - 3 mol = 2 mol

Therefore, the value of Δng for the combustion of 1 mol of liquid ethanol is 2 mol. This means that during the reaction, the number of moles of gaseous substances increases by 2 mol.

**Summary**
The value of Δng for the combustion of 1 mol of liquid ethanol is 2 mol. This indicates that the reaction produces 2 moles of gaseous products (carbon dioxide and water vapor) for each mole of gaseous reactant (oxygen) consumed.

1 mole ethanol liquid react with 3 mole oxygen gas to form 2 mol co2 and 3 mole water liquid.(by balanced equation)
delta ng .:: mole of gaseous product minus mole of gaseous reactant
hence 2 - 3 ::::: -1