1. Given data

• External pressure on the movable piston: $P_{\text{ext}} = 1\,\text{atm}$
• Temperature: $T = 0^\circ C = 273\,\text{K}$
• Helium added: $n_{\text{He}} = 0.1\,\text{mol}$
• Unknown compound added: $n_X = 1.0\,\text{mol}$
• Vapour pressure of unknown compound at $0^\circ C$: $P_X^* = 0.68\,\text{atm}$

We are asked to find the total volume of gases in the vessel.

2. Key idea: movable piston with external pressure $1$ atm

Since the piston is movable, at equilibrium the total pressure inside becomes:

$P_{\text{total}} = 1\,\text{atm}$

Now check the unknown compound.

Its vapour pressure at $0^\circ C$ is only $0.68\,\text{atm}$. Therefore, at equilibrium, the partial pressure of its vapour cannot exceed $0.68\,\text{atm}$.

So the unknown compound will exist as vapour + condensed phase, with

$P_X = 0.68\,\text{atm}$

The remaining pressure must be due to helium:

$P_{\text{He}} = 1 - 0.68 = 0.32\,\text{atm}$

3. Use ideal gas law for helium

For helium,

$P_{\text{He}}V = n_{\text{He}}RT$

Using $R = 0.0821\,\text{L atm mol}^{-1}\text{K}^{-1}$,

$V = \frac{n_{\text{He}}RT}{P_{\text{He}}}$

$V = \frac{(0.1)(0.0821)(273)}{0.32}$

First compute numerator:

$0.0821 \times 273 \approx 22.41$

So,

$V = \frac{0.1 \times 22.41}{0.32} = \frac{2.241}{0.32} \approx 7.00\,\text{L}$

4. Check whether 1.0 mol of unknown compound is enough to maintain $0.68$ atm

If the unknown vapour is at $0.68$ atm in volume $7.0$ L, moles in vapour phase would be:

$n_{X,\text{vap}} = \frac{P_XV}{RT} = \frac{(0.68)(7.0)}{22.41}$

$n_{X,\text{vap}} \approx 0.212\,\text{mol}$

Since total amount given is $1.0$ mol, this is sufficient; the rest condenses.

Thus our assumption is valid.

5. Final answer

$\boxed{7\,\text{L}}$

So the total volume is close to $7$ litre.