20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$$^{-1}$$)

The precipitation reaction involved is:
NaI + AgNO3 → AgI ↓ + NaNO3

The stoichiometry is 1 : 1, so the moles of NaI present in the sample equal the moles of AgI obtained.

Molar mass of silver iodide (AgI):
Ag = 108 g mol$$^{-1}$$, I = 127 g mol$$^{-1}$$
Therefore, molar mass of AgI = $$108 + 127 = 235$$ g mol$$^{-1}$$.

Moles of AgI formed:
$$n = \frac{4.74}{235} = 0.0202$$ mol

Because of the 1 : 1 ratio, moles of NaI in 20 mL solution = $$0.0202$$ mol.

Volume of the NaI solution = 20 mL = 0.020 L.

Molarity, $$M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.0202}{0.020} = 1.01$$ M.

Rounded to two significant figures, the molarity is approximately $$1$$ M.

Answer : 1 M