Step 1: Understanding the Question:
The problem asks us to find a pair of solutions that will not exhibit net flow of solvent in either direction when separated by a semipermeable membrane. This condition occurs when the two solutions have identical osmotic pressures ($\pi$), meaning they are isotonic.
Step 2: Key Formula or Approach:
Osmotic pressure is given by the formula $\pi = iCRT$. Since both urea and glucose are non-electrolytes, their van 't Hoff factor is $i = 1$. Thus, for the osmotic pressures to be equal at a given temperature, their molar concentrations ($C = \frac{n}{V}$) must be equal. Given that concentrations are expressed per $\text{dm}^3$ ($V = 1\ \text{dm}^3$), we need to find the option where the number of moles of urea equals the number of moles of glucose:
$$ n = \frac{\text{Mass}}{\text{Molar Mass}} $$
Step 3: Detailed Explanation:
Let's calculate the moles for option (D):
For urea:
$$ n_{\text{urea}} = \frac{12\ \text{g}}{60\ \text{g/mol}} = 0.2\ \text{mol} $$
For glucose:
$$ n_{\text{glucose}} = \frac{36\ \text{g}}{180\ \text{g/mol}} = 0.2\ \text{mol} $$
Since $n_{\text{urea}} = n_{\text{glucose}}$ in a volume of $1\ \text{dm}^3$, their concentrations are identical ($0.2\ \text{M}$). Consequently, their osmotic pressures are equal ($\pi_{\text{urea}} = \pi_{\text{glucose}}$), resulting in no net movement of solvent across the semipermeable membrane.
Step 4: Final Answer:
The solutions that do not flow in either direction are 12 g urea $\text{dm}^{-3}$ and 36 g glucose $\text{dm}^{-3}$, corresponding to option (D).