1. Energy of one photon

The energy of a photon is given by:

$E = h\nu$

Given:

$h = 6.626 \times 10^{-34}\ \text{J s}$ $\nu = 2 \times 10^{12}\ \text{Hz}$

So,

$E = (6.626 \times 10^{-34})(2 \times 10^{12})$

$E = 13.252 \times 10^{-22}\ \text{J}$

$E = 1.3252 \times 10^{-21}\ \text{J per photon}$

2. Energy of one mole of photons

One mole contains Avogadro's number of photons:

$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$

Therefore,

$E_{\text{mol}} = E \times N_A$

$E_{\text{mol}} = (1.3252 \times 10^{-21})(6.022 \times 10^{23})$

Now multiply the coefficients:

$1.3252 \times 6.022 = 7.9807544$

And combine powers of 10:

$10^{-21} \times 10^{23} = 10^2$

Thus,

$E_{\text{mol}} = 7.9807544 \times 10^2\ \text{J mol}^{-1}$

$E_{\text{mol}} = 798.07544\ \text{J mol}^{-1}$

3. Nearest integer

$E_{\text{mol}} \approx 798\ \text{J mol}^{-1}$

Final Answer

$\boxed{798}$

The derived answer matches the stored correct answer.