Question

1. The mass of [Al(H₂O)₆]Cl₃ needed to prepare 25.0 mL of a 0.10 M solution (Part I of this experiment) is

0.265g

0.21g

0.1338g

0.6038g

2.

The mass of NaHCO₃ needed to prepare 25.0 mL of a 0.10 M solution (Part I of this experiment) is:

		0.265g
		0.21g
		0.1338g
		0.6038g

Answer 1

1)

volume , V = 25.0 mL = 2.5*10^-2 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.1*0.025

= 2.5*10^-3 mol

Molar mass of Al(H2O)6Cl3 = 1*MM(Al) + 12*MM(H) + 6*MM(O) + 3*MM(Cl)

= 1*26.98 + 12*1.008 + 6*16.0 + 3*35.45

= 241.426 g/mol

we have below equation to be used:

mass of Al(H2O)6Cl3,

m = number of mol * molar mass

= 2.5*10^-3 mol * 241.426 g/mol

= 0.6038 g

Answer: 0.6038 g

2)

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.1*0.025

= 2.5*10^-3 mol

Molar mass of NaHCO3 = 1*MM(Na) + 1*MM(H) + 1*MM(C) + 3*MM(O)

= 1*22.99 + 1*1.008 + 1*12.01 + 3*16.0

= 84.008 g/mol

we have below equation to be used:

mass of NaHCO3,

m = number of mol * molar mass

= 2.5*10^-3 mol * 84.008 g/mol

= 0.21 g

Answer: 0.21 g