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)
Answer
1
Solution
NaI(aq) + AgNO3(aq) AgI(s) + NaNO3(aq)
Molar mass of AgI = 108 + 127 = 235 g mol-1
nAgI = 4.74235 = 0.0202 mol
1 mol NaI gives 1 mol AgI nNaI = 0.0202 mol
Molarity = 0.0202 mol0.020 L = 1.01 M 1 M
Answer: 1