Question

Question #54ef3

Answer 1

Determine moles of AgI formed, that is same as number of moles of AgNO3, then divide moles by 0.050 L

Explanation:

AgNO3(aq) + HI(g) --> AgI (s) + HNO3(aq)

1 mole of silver nitrate yields 1 mole of silver iodide
molar mass of Ag I = 234.77g/mol

mole AgI `= (2.35g)( (1mol)/(234.77g))`

`mol AgI = 0.0100 mol`

there is one mole of Ag in each mole of AgI so there was 0.0100 mol of silver in the original 50.0 mL solution

Molarity is mol/L

Molarity `= (0.0100mol)/(0.050L)`

Molarity `= 0.20 M`