Answer to Question #56488 in General Chemistry for Samantha

Question #56488
mass of glucose needed to prepare 215.0 mL of 19% w/v glucose (C6H12O6) express answer using 2 significant figures.
Expert's answer
W = 19 %
V = 215.0 mL
W = m*100 %/V
m = W*V/100 % = 19 %*215.0 mL/100 % = 40.85 g ≈ 41 g

Answer:
41 g

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