The rate constant of a first order reaction is $1.15 \times 10^{-3} \mathrm{~s}^{-1}$. How long will 5 g of reactant take to reduce to 3 g ?

The integrated rate law for a first-order reaction is: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.

Given $k = 1.15 \times 10^{-3} \, \text{s}^{-1}$, $[A]_0 = 5 \, \text{g}$, and $[A]_t = 3 \, \text{g}$, substitute into the formula to obtain:
$t = \frac{2.303}{1.15 \times 10^{-3}} \log \left(\frac{5}{3}\right)$.

The ratio $\frac{5}{3} \approx 1.6667$, and $\log(1.6667) \approx 0.2218$. With this, the expression becomes:
$t = \frac{2.303 \cdot 0.2218}{1.15 \times 10^{-3}} = \frac{0.5108}{0.00115} \approx 444 \, \text{s}$.

Matching the result with the options, the correct choice is D.