Question

A student placed 13.0 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 60.0-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

m_{C_6H_{12}O_6}=1.56g

Step-by-step explanation:

Hello,

After the dilution, the concentration of glucose is:

`c=(13.0g)/(100mL)=0.13g/mL`

Now, we apply the dilution equation to compute the concentration after the dilution to 0.500L (500mL):

`c_1*V_1=c_2*V_2\\c_2=(c_1*V_1)/(V_2)=(0.13g/mL*60.0mL)/(500mL) \\c_2=0.0156g/mL`

Finally, we compute the present grams in 100mL of the 0.0156g/mL glucose solution as shown below:

`m_{C_6H_(12)O_6} =100mL*0.0156g/mL\\m_{C_6H_(12)O_6}=1.56g`

Best regards.

Answer 2

There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.

Step-by-step explanation:

13.0 grams of glucose was added to volumetric flask and solution was made up to 100 mL.

Moles of glucose =
`(13.0 g)/(180.156 g/mol)=0.07215 mol`

Volume of the solution = 100 mL = 0.1 L

Molarity of the solution =
`\frac{Moles}{\text{Volume of solution}}`

`M=(0.07215 mol)/(0.1 L)=0.7215 M`

60 mL of 0.7215 M was diluted to 0.500 L, the molarity of the solution after dilution be
`M_2`

`M_1=0.7215 M,V_2=60 mL=0.060 L`

`M_2=?,V_2=0.500 L`

`M_1V_1=M_2V_2` (Dilution)

`M_2=(0.7215 M* 0.060 L)/(0.500 L)`

`M_2=0.08658 M`

Mass of glucose in 0.08658 M glucose solution:

In 1 L of solution = 0.08658 moles

In 1000 mL of solution = 0.08658 moles

Then in 100 mL of solution :

`(0.08658)/(1000)* 100 mol=0.008658 mol` of glucose

Mass of 0.008658 moles of glucose:

`0.008658 mol* 180.156 g/mol=1.5597 g`

There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.