The solubility of $$N_2$$ in water at 300 K and 500 torr partial pressure is 0.01 g L$$^{-1}$$. The solubility (in g L$$^{-1}$$) at 750 torr partial pressure is:

We are asked to compare the solubility of molecular nitrogen, $$N_2$$, in water at two different partial pressures while the temperature is kept constant at 300 K. The linking principle here is Henry’s law, which tells us that, at a fixed temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the solution.

First, we explicitly state Henry’s law in its simplest proportional form:

$$ S \; \propto \; P $$

where $$S$$ is the solubility (in g L$$^{-1}$$ here) and $$P$$ is the partial pressure (in torr here). Whenever two sets of conditions are compared at the same temperature, we can write the proportionality as an equality of ratios:

$$ \frac{S_1}{P_1} \;=\; \frac{S_2}{P_2}. $$

Now we identify the known quantities from the problem statement. At the first set of conditions we have

$$ S_1 = 0.01\; \text{g L}^{-1}, \qquad P_1 = 500\; \text{torr}. $$

At the second set of conditions we know the new pressure but not the solubility:

$$ P_2 = 750\; \text{torr}, \qquad S_2 = ? $$

Substituting these numerical values into the Henry’s-law ratio gives

$$ \frac{0.01}{500} \;=\; \frac{S_2}{750}. $$

To isolate $$S_2$$, we multiply both sides of the equation by 750 torr:

$$ S_2 \;=\; 750 \times \frac{0.01}{500}. $$

Now we carry out the arithmetic step by step. First divide 750 by 500:

$$ \frac{750}{500} \;=\; 1.5. $$

Then multiply this factor by 0.01 g L$$^{-1}$$:

$$ S_2 \;=\; 1.5 \times 0.01 \;=\; 0.015. $$

So the solubility of $$N_2$$ in water at 750 torr and 300 K is

$$ S_2 = 0.015\; \text{g L}^{-1}. $$

Scanning the options provided, 0.015 g L$$^{-1}$$ corresponds to Option D.

Hence, the correct answer is Option D.