Question

You need to make an aqueous solution of 0.230 M aluminum nitrate for an experiment in a lab. Using a 250 ml volumetric flask how much solid aluminum nitrate should you add

Answer 1

12.25 g of aluminium nitrate Al(NO₃)₃ are needed to prepare the solution.

Explanation:

First we determine the number of moles of aluminium nitrate:

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of aluminium nitrate = 0.230 × 0.250

number of moles of aluminium nitrate = 0.0575 moles

Then we calculate the mass of aluminium nitrate Al(NO₃)₃:

mass = number of moles × molecular weight

mass of aluminium nitrate = 0.0575 × 213

mass of aluminium nitrate = 12.25 g

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