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de Edit View History Bookmarks People Window Help Am x Mic x s 020 x Pre x M inbox Virg x D Get x Mic x D Mic x Secure https://scholar.vt.edu/access/content/group/003e60a8-8bdc-46c3-9a3b-4089a6e551ef/Lectures/o20117.pd Molarity samples How many grams of glucose (MW 180.15) would be required to prepare 500 mL of 50 mM solution? A) 5 g 180.15 g/mol x 0.5 Lx 0.05 mol/L B) 4.5 mg C) 4.5 g 4.5 g D) 45 g
I don't understand why they did what they did. The answer is in red.
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Answer #1

Molairty = (weight / mol.weight) X 1 / V (litre)

You need to find out weight of glucose?

They have given everything M = 0.05 mol/l, mol. weight = 180, V = 0.5 L

Weight = molarity X mol. weight X Volume

If we substitute, we get weight of glucose

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