1. Identify the type of deviation from Raoult’s law

For an ideal solution, $P_{\text{total}}=x_{CS_2}P^0_{CS_2}+x_{\text{acetone}}P^0_{\text{acetone}}$ so the total vapour pressure must lie between the vapour pressures of the two pure liquids.

Given: $P^0_{CS_2}=512\ \text{mm Hg},\qquad P^0_{\text{acetone}}=344\ \text{mm Hg}$

But for the solution, $P_{\text{total}}=600\ \text{mm Hg}$

Since $600>512>344$ it is greater than even the vapour pressure of pure $CS_2$.

Hence the solution shows positive deviation from Raoult’s law.

2. Interpretation of positive deviation

Positive deviation means:

• unlike intermolecular attractions ($CS_2$–acetone) are weaker than like attractions ($CS_2$–$CS_2$ and acetone–acetone),
• molecules escape more easily into vapour phase,
• mixing is generally endothermic ($\Delta H_{mix}>0$),
• volume of mixing generally shows expansion ($\Delta V_{mix}>0$).

3. Check each option

Option A

$CS_2$ and acetone are less attracted to each other than to themselves.

This is exactly the condition for positive deviation.

So, A is true.

Option B

A mixture of 100 ml $CS_2$ and 100 ml acetone has a volume $<200$ ml.

For positive deviation, intermolecular attraction between unlike molecules is weaker, so mixing tends to cause expansion, i.e. $V_{mix}>0$ Therefore total volume should be greater than 200 ml, not less.

So, B is false.

Option C

Heat must be absorbed in order to produce the solution at $35^\circ C$.

Positive deviation implies mixing is endothermic: $\Delta H_{mix}>0$ So heat is absorbed.

Thus, C is true.

Option D

Raoult’s law is not obeyed by this system.

Since the solution shows positive deviation, it does not obey Raoult’s law.

So, D is true.

4. Conclusion

The false statement is: $\boxed{\text{B}}$