What volume of each solution contains 0.220mol of NaI?
a.) 0.142M NaI
b.) 0.902M NaI
c.) 1.61M NaI
To determine the volume of each solution that contains 0.220 mol of NaI, we can use the formula:
Volume = (moles of solute) / (molarity)
Let's calculate the volume for each solution:
a) 0.142M NaI:
Volume = 0.220 mol / 0.142 mol/L = 1.55 L
b) 0.902M NaI:
Volume = 0.220 mol / 0.902 mol/L = 0.244 L
c) 1.61M NaI:
Volume = 0.220 mol / 1.61 mol/L = 0.137 L
Therefore, the volume of each solution that contains 0.220 mol of NaI is:
a) 1.55 L
b) 0.244 L
c) 0.137 L
To find the volume of each solution that contains 0.220 mol of NaI, we can use the formula:
Volume (in Liters) = moles of solute / molarity
a.) For a 0.142 M NaI solution:
Volume = 0.220 mol / 0.142 mol/L
Volume = 1.55 L
b.) For a 0.902 M NaI solution:
Volume = 0.220 mol / 0.902 mol/L
Volume = 0.244 L
c.) For a 1.61 M NaI solution:
Volume = 0.220 mol / 1.61 mol/L
Volume = 0.137 L
So, the volume of each solution that contains 0.220 mol of NaI is:
a.) 1.55 L
b.) 0.244 L
c.) 0.137 L
M = mols/L.
0.142M NaI = 0.220 mols NaI/ L solution.
Solve for L.