Chemistry Tutor: None Selected Time limit: 1 Day

1) The solubility of solid R in water is: 0.71 g/100 mL at 0°C, 24.0g/100 mL at 100°C. a) How many mL of boiling water are required to dissolve 94.5 g of R?(report to the nearest mL) If solution were cooled to 0°C, how many grams of R would crystallize out? (report to one decimal place) (Show calculations)

Oct 3rd, 2015

A) Set up proportion calculation:

24.0g/100 mL = 94.5g / ? mL (cross multiply)

(94.5 / 24) x 100 = 393.75 mL  (or 393.8 mL) of boiling water required to dissolve 94.5 g of R

B) You will need to determine how much of R will be soluble at 0 degrees in 393.75 mL of water. Set up a proportion again.

0.71 g / 100 mL = ? g/ 393.75 mL

(393.75 x 0.71) / 100 = 2.795625 = around 2.8 grams would crystallize.

Please let me know if you need any clarification. I'm always happy to answer your questions. Thank you very much! Please let me know if you need any clarification. I'm always happy to answer your questions. Thank you very much!
Sep 19th, 2015

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Oct 3rd, 2015
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Oct 3rd, 2015
Dec 7th, 2016
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