Question

What mass, in grams, of aluminum hydroxide will be required to prepare the 4 L of a 1.75 M solution?

Answer 1

Answer: 0.53g

Explanation:

No of moles= volume ×molarity/1000

We have the volume and the molarity

Volume=4L

Molarity=1.7M

No of moles = 4×1.7/1000

No of moles= 0.0068moles

Remember also that

No of moles= mass given/molar mass

Molar mass of Al(OH)3

Al= 27

O=16

H=1

Molar mass = Al+(O+H)3

Molar mass= 27+(16+1)3

Molar mass= 27+(17)3

Molar mass = 27+51

Molar mass= 78g/mol

To get the mass

Mass given = no of moles × molar mass

Mass= 0.0068×78

Mass= 0.53g