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Pad 3:21 PM * 30% LJ sjc.cengagenow.com instructure-uploads.s3.amazonaws.com/account_11.. An Aqueous Solution Of Potassium Acetate,, Contai... OWLv2 Online teaching and learning resource fr... 1. Types of Solutions 2. Solubility of Ionic Compounds 3. Molarity Use the References to access important values if needed for this question. ou wish to prepare 268 grams of 34.3% (w/w) Fe2 (SO4)3 . Assume that the density pts ? 1 pts 2req 1 pts 2rec 1 pts 2req of water is 1.00 g/mL. You will need grams of iron (III) sulfate and mL of water 4. Percent Concentration: Weight it Answer Retry Entire Group 4 more group attempts remaining 5, % (w/w) Calculations: Mass of So In progress %(w/w) Calculations: Mass of Solute or Volume of Solvent Required Question 6. Dilution 1 pts 2req 7. Solution Stoichiometry: Volume . 1 pts 2req 8. Acids in Water 1 pts 2req 1 pts 2req pts 2req 9. Review: Molarity 10. Titration: Molarity of Acid/Base Progress 2/10 groups Due Apr 22 at 11:55 PM Pre Next Finish Assianment Save and Fxit

Explanation / Answer

% (w/w) = mass of Fe2(SO4)3 x 100 / mass of solution

34.3 = mass of Fe2(SO4)3 x 100 / 268 g

mass of Fe2(SO4)3 = 91.924 g

now

mass of water = 268 g - 91.924 g

mass of water = 176.076 g

now

volume of water = 176.076 g / (1 g/ml) = 176.076 ml

so

you will need 91.924 grams of iron (iii) sulfate and 176.076 ml of water

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