In part A I found the mass of sucrose to be 1.82 g however I don’t know if that’

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In part A I found the mass of sucrose to be 1.82 g however I don’t know if that’s relevant here or not
In part A I found the mass of sucrose to be 1.82 g however I don’t know if that’s relevant here or not
In part A I found the mass of sucrose to be 1.82 g however I don’t know if that’s relevant here or not NAME: EXPERIMENT 17: Concentration Units DATA SHEET 2 D. Percent by Mass of Solution LAB SECTION: Assigned Mass of Solution Assigned % by Mass Mass of Sucrose Mass of Water 20 /- m/m) E. Preparation of a Molal Solutiorn 40.49a 0-0442 Assigned Mass of Water Assigned Molality Mass of Sucrose Mass of Prepared Solution

Explanation / Answer

D. Percent mass is grams solute / grams of solution so

(mass of solute / 75 g) x 100 = 20

mass of solute = 20 * 75 / 100 = 15 grams of solute (sucrose)

total mass solution = 75

15 + x = 75

x = 75 - 15 = 60 grams of water

E.

molality is moles / kg of solvent

water is the solvent you have 90.49 grams

divide this by 1000 to get 0.09049 Kg

moles = molality * kg solvent = 0.0442 * 0.09049 = 0.004 moles of sucrose

mass = moles * molar mass , molar mass of sucrose 342.296g/mol

mass = 0.004 * 342.296 = 1.369 gramss of sucrose

Mass of sucrose is 1.369 grams

mass of prepared solution = 90.49 + 1.369 = 91.859 grams

*i dont know where does the value 1.82 comes from, hope this helps =)

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