Question

How many g of AlCl3 are in 63.1 mL of 4.20 M solution?

(Give correct sig figs in units of moles without scientific notation)

Answer 1

The first step is to convert mL to L.

63.1 mL * (1 L / 1000 mL) = 0.0631 L

Next, we can use the formula:

moles = M * L

moles of AlCl3 = 4.20 M * 0.0631 L = 0.26502 moles

Finally, we can use the molar mass of AlCl3 to convert moles to grams:

0.26502 moles * 133.34 g/mol = 35.34 g

Rounded to the correct number of significant figures, the answer is:

35 g of AlCl3