Answer:
volume in L = 0.25 L
Explanation:
Given data:
Mass of Cu(NO₃)₂ = 2.43 g
Volume of KI = ?
Solution:
Balanced chemical equation:
2Cu(NO₃)₂ + 4KI → 2CuI + I₂ + 4KNO₃
Moles of Cu(NO₃)₂:
Number of moles = mass/ molar mass
Number of moles = 2.43 g/ 187.56 g/mol
Number of moles = 0.013 mol
Now we will compare the moles of Cu(NO₃)₂ with KI.
Cu(NO₃)₂ : KI
2 : 4
0.013 : 4 × 0.013=0.052 mol
Volume of KI:
<em>Molarity = moles of solute / volume in L</em>
volume in L = moles of solute /Molarity
volume in L = 0.052 mol / 0.209 mol/L
volume in L = 0.25 L
