Question

A student placed 18.5 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 30.0 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?

Answer 1

1.30464 grams of glucose was present in 100.0 mL of final solution.

Step-by-step explanation:

`Molarity=\frac{moles}{\text{Volume of solution(L)}}`

Moles of glucose =
`(18.5 g)/(180 g/mol)=0.1028 mol`

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution =
`(0.1028 mol)/(0.1 L)=1.028 mol/L`

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution =
`M_1=1.208 mol`

Volume of the solution taken =
`V_1=30.0 mL`

Molarity of the solution after dilution =
`M_2`

Volume of the solution after dilution=
`V_2=0.500L = 500 mL`

`M_1V_1=M_2V_2`

`M_2=(M_1V_1)/(V_2)=(1.208 mol/L* 30.0 mL)/(500 mL)`

`M_2=0.07248 mol/L`

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

`0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}`

Moles of glucose =
`0.07248 mol/L* 0.1 L=0.007248 mol`

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.