Question

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. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)

Hint:

- In precipitation reactions, stoichiometry is 1:1 for simple salts - Always convert mass to moles using molar mass - Remember to convert mL to L for molarity calculations - Nearest integer rounding for final answer

Answer 1

Correct Answer: 1

The molarity of the sodium iodide (NaI) solution is determined by the following procedure:

1. Reaction Equation:

Sodium iodide (NaI) reacts with silver nitrate (AgNO₃) to yield silver iodide (AgI) and sodium nitrate (NaNO₃):

NaI + AgNO₃ → AgI + NaNO₃

2. Moles of Silver Iodide Calculation:

The molar mass of AgI is calculated as the sum of the atomic masses of silver (Ag) and iodine (I): 108 g/mol + 127 g/mol = 235 g/mol. With a given mass of silver iodide of 4.74 g:

Moles of AgI = 4.74 g / 235 g/mol = 0.02017 mol

3. Mole Ratio to Sodium Iodide:

According to the balanced equation, the moles of NaI are stoichiometrically equivalent to the moles of AgI:

Moles of NaI = 0.02017 mol

4. Molarity of NaI Calculation:

The volume of the NaI solution is 20 mL, which is converted to liters: 0.020 L.

Molarity (M) = Moles of solute / Volume of solution in liters = 0.02017 mol / 0.020 L = 1.0085 M

Result:

The molarity of the sodium iodide solution, rounded to the nearest integer, is 1 M. This value falls within the expected range of 1 to 1, confirming the molarity as 1 M.