42.12% (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol is
[Given: Molar mass: Na = 23.0 g mol$$^{-1}$$; Cl = 35.5 g mol$$^{-1}$$]

Given:

$$42.12\% \text{ (w/v) solution of NaCl}$$

This means:

$$42.12\,\text{g NaCl present in }100\,\text{mL solution}$$

Molar mass of NaCl:

$$=23+35.5$$

$$=58.5\,\text{g mol}^{-1}$$

Number of moles of NaCl:

$$=\frac{42.12}{58.5}$$

$$=0.72\,\text{mol}$$

Volume of solution:

$$100\,\text{mL}=0.1\,\text{L}$$

Hence molarity of NaCl solution:

$$=\frac{0.72}{0.1}$$

$$=7.2\,\text{mol L}^{-1}$$

The solution causes coagulation in $$10$$ hours.

Coagulating value is expressed in millimoles per litre:

$$\text{Coagulating value}=\frac{7.2 \times 1000}{5}$$

$$=1440\,\text{mmol L}^{-1}$$

Final Answer:

$${1440\,\text{mmol L}^{-1}}$$

Hence, Option D is correct