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

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