Answer to Question #222968 in Inorganic Chemistry for SAmanta Ariel

Question #222968

What is the mass of Al2(SO4)3 needed to prepare 40 (p3)ml of (0.2p2)M of sulphate ions in aluminum sulphate solution? where p= log(moles of Al2(SO4)3 ?


1
Expert's answer
2021-08-09T07:47:23-0400

"Al_2(SO_4)_3" =342.15 g/mol


"Moles =volume\u00d7Molarity"


"Moles =0.04\u00d70.2=0.008moles"


"Mass =0.008\u00d7342.15=2.74grams"



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