Let's solve the problem step-by-step considering the correct rules for significant digits.
Step 1: Find the molar mass of AlCl₃.
Aluminum (Al) = 26.98 g/mol
Chlorine (Cl) = 35.45 g/mol
There are three chlorine atoms in aluminum chloride.
Molar mass of AlCl₃ = 26.98 + 3(35.45)
= 26.98 + 106.35
= 133.33 g/mol
Step 2: Calculate the number of moles of AlCl₃.
Given:
Mass of AlCl₃ = 125 g
Molar mass of AlCl₃ = 133.33 g/mol
Number of moles (n) = Mass / Molar mass
[tex]\( n = \frac{125\text{ g}}{133.33\text{ g/mol}} \approx 0.9375 \)[/tex] moles
Step 3: Calculate the volume of the solution using the molarity.
Given:
Molarity (M) = 0.135 mol/L
[tex]\( Molarity = \frac{Number\ of\ moles}{Volume\ in\ liters} \)[/tex]
[tex]\( Volume\ in\ liters = \frac{Number\ of\ moles}{Molarity} \)[/tex]
[tex]\( Volume\ in\ liters = \frac{0.9375\text{ moles}}{0.135\text{ mol/L}} \approx 6.9446\text{ L} \)[/tex]
Step 4: Apply significant digits.
The least number of significant digits in the given data is 3 (for the molarity 0.135).
Therefore, the volume should be reported with 3 significant digits:
[tex]\( Volume\ in\ liters \approx 6.94\text{ L} \)[/tex]
So, the volume in liters of a 0.135 mol/L aluminum chloride solution containing 125 g of aluminum chloride is approximately 6.94 liters.
