Analytical Chemistry

The key to determining the number of significant figures in a number is to identify all of the meaningful digits. In the given number, , we focus on the decimal number .

1. **Identify Non-Zero Digits**: All non-zero digits are inherently significant. Here, the digits 5, 8, and 2 are non-zero.

2. **Zeroes**: Zeroes can be tricky, but in this context, any zeroes between non-zero digits or following a non-zero digit after a decimal point are significant.

In , both zeroes are significant: the one between 8 and 2, and the trailing zero after the 2, since it indicates precision.

Thus, we have five significant figures: 5, 0, 8, 2, and the final 0.

Therefore, the number of significant figures in is 5.

To determine the weight of the precipitate formed from a 0.1000 g sample of AlCl₃ when precipitated with 8-hydroxyquinoline, we need to follow these steps:

Step 1: Determine moles of AlCl₃.

The molecular weight of AlCl₃ is given as 133.34 g/mol. The moles of AlCl₃ can be calculated as follows:

moles of AlCl₃=mass of AlCl₃molecular weight of AlCl₃=0.1000 g133.34 g/mol=0.000750 mol

Step 2: Relate Al in AlCl₃ to AlC₉H₆NO (precipitate).

The balanced equation for the formation of the aluminum-8-hydroxyquinoline precipitate (Al(C₉H₆ON)₃) is:

AlCl₃ + 3C₉H₆NO → Al(C₉H₆ON)₃ + 3HCl

One mole of AlCl₃ yields one mole of Al(C₉H₆ON)₃ as the precipitate. Hence, moles of AlCl₃ = moles of precipitate formed.

Step 3: Calculate weight of the precipitate.

The molecular weight of Al(C₉H₆ON)₃ is calculated as 26.98 + (3×145.16) = 462.46 g/mol.

The weight of the precipitate is calculated as follows:

Weight of precipitate=moles of precipitate×molecular weight of precipitate=0.000750 mol×462.46 g/mol=0.3468 g

Conclusion: The weight of the precipitate is 0.3468 g.