A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water = $$2.10 \times 10^{6}$$ J kg$$^{-1}$$ and Latent heat of Fusion of water = $$3.36 \times 10^{5}$$ J kg$$^{-1}$$)

We begin with energy conservation because the vessel is thermally insulated. Whatever heat is absorbed to evaporate some water must be exactly supplied by the heat released when another part of the water freezes. No heat can enter or leave the vessel.

Let

$$m \text{ g}$$ be the mass of water that evaporates,

$$x \text{ g}$$ be the mass of water that freezes into ice.

The latent heat values given are

$$L_v = 2.10 \times 10^{6}\ \text{J kg}^{-1}=2.10 \times 10^{3}\ \text{J g}^{-1},$$

$$L_f = 3.36 \times 10^{5}\ \text{J kg}^{-1}=3.36 \times 10^{2}\ \text{J g}^{-1}.$$

First we write the formulae for the heats involved:

Heat absorbed in vaporising $$m$$ grams of water at 0 °C:

$$Q_{\text{vap}} = m\,L_v.$$

Heat released when $$x$$ grams of water freeze at 0 °C:

$$Q_{\text{freeze}} = x\,L_f.$$

Because the vessel is adiabatic, these two quantities must be equal:

$$Q_{\text{vap}} = Q_{\text{freeze}}.$$

So we have

$$m\,L_v = x\,L_f.$$

Substituting the numerical values of the latent heats:

$$m\,(2.10 \times 10^{3}) = x\,(3.36 \times 10^{2}).$$

Now we isolate $$x$$ in terms of $$m$$:

$$x = \frac{2.10 \times 10^{3}}{3.36 \times 10^{2}}\,m.$$

Evaluating the fraction,

$$\frac{2.10 \times 10^{3}}{3.36 \times 10^{2}}=\frac{2100}{336}\approx 6.25.$$

Hence

$$x \approx 6.25\,m.$$

Next we use the fact that the total mass of water present initially was 150 g. After the process finishes we have evaporated mass $$m$$, frozen mass $$x$$, and whatever remains as liquid. Therefore the sum of the masses that disappeared from the liquid phase must not exceed 150 g:

$$x + m \le 150.$$

Substituting $$x = 6.25\,m$$ into this inequality gives

$$6.25\,m + m \le 150,$$

$$7.25\,m \le 150.$$

Solving for $$m$$,

$$m \le \frac{150}{7.25}.$$

Carrying out the division,

$$\frac{150}{7.25} \approx 20.7\ \text{g}.$$

The amount that actually evaporates will be this maximum value because once all available heat from freezing has been used, no further evaporation can occur while still maintaining 0 °C. Thus the evaporated mass is about 20 g, which matches option B most closely.

Hence, the correct answer is Option B.