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A student placed 19.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to...

A student placed 19.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 50.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?

Express your answer to three significant figures and include the appropriate units.

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Answer #1

No of moles of C6H12O6 solute
Volume of solution = 100 mL = 0.100L
Molar concentration of C6H1206 = no of moles of C6H12O6 / Volume of soln in liters
= 0.105 mol / 0.100L
= 1.05 mol / L = 1.05 M
Concentration of the stock C6H1206 solution,
Volume of the stock C6Hn06 solution, VI = mL
Volume of the dilute C6H1206 solution, = L =
Concentration of the dilute C6H1206 solution, M
According to dilution principle,
Concentration of dilute C6H1206 solution,
Concentration of the dilute C6H1206 solution,
Volume of solution = 100 mL = 0.100L,
Amount of C6H1206 solute present =

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