7-2:
1% w/v means 1 gram of solute dissolved in 100 mL of solvent, 1 g = 1000 mg
90 mg/100mL = 0.09 g / 100 mL = 0.09%
w/v % concentration = 0.09%
molarity of glucose = [mass of glucose / molar mass of glucose] x [1000 / volume in mL]
molarity of glucose = [0.09g / 180.2 g/mol] x [1000 / 100 mL]
molarity of glucose = 0.005 mol/L = 0.05 M
molarity of glucose = 0.05 M
7-3:
1.85 % w/v means 1.85 g dissolved in 100 mL = 1.85g/100 mL = 0.0185 g / mL
750 mL ==> 750. mL x 0.0185 g/mL = 13.875 g
Transfer 13.875 g of H3BO3 into a clean, dry graduated 1000 mL measuring cylinder and add distilled, de-ionized pure water upto the 750 mL mark.
Hope this helped you!
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= * Pau CHON + NO-022 0 Danielle Mitchell Homework Lecture 7 Name Danielle Min _Morning...