Part B A student placed 19.5 g of glucose (C6H12O6) in a volumetric flask, added

Question

Part B A student placed 19.5 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 40.0 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.

Explanation / Answer

we know that

moles = mass / molar mass

molar mass of glucose = 180 g /mol

so

initial moles of glucose = 19.5 / 180 = 0.1083333

now

water is added upto the 100 ml mark

we know that

concentration = moles x 1000 / volume (ml)

so

initial concentration of glucose = 0.1083333 x 1000 / 100

initial concentration of glucose = 1.083333 M

now

40 ml sample is taken from the initial glucose solution

now

moles of glucose taken = concentration x volume (ml) / 1000

moles of glucose taken = 1.08333 x 40 / 1000

moles of glucose taken = 0.0433333

now

it is diluted too 0.5 L

so

final concentration of the sample = moles of glucose / volume (L)

final concentration of the sample = 0.04333 / 0.5

final concentration of the sample = 0.0866666

now

100 ml of the final sample is taken

moles of glucose present = final concentration x volume (ml) / 1000

so

moles of glucose present = 0.08666 x 100 / 1000

moles of glucose present = 8.66667 x 10-3

now

we know that

mass = moles x molar mass

so

mass of glucose present = 8.6667 x 10-3 x 180

mass of glucose present = 1.56 g

so

1.56 grams of glucose is present in 100 ml of the final solution

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