Answer:
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Explanation:
13.0 grams of glucose was added to volumetric flask and solution was made up to 100 mL.
Moles of glucose = [tex]\frac{13.0 g}{180.156 g/mol}=0.07215 mol[/tex]
Volume of the solution = 100 mL = 0.1 L
Molarity of the solution = [tex]\frac{Moles}{\text{Volume of solution}}[/tex]
[tex]M=\frac{0.07215 mol}{0.1 L}=0.7215 M[/tex]
60 mL of 0.7215 M was diluted to 0.500 L, the molarity of the solution after dilution be [tex]M_2[/tex]
[tex]M_1=0.7215 M,V_2=60 mL=0.060 L[/tex]
[tex]M_2=?,V_2=0.500 L[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution)
[tex]M_2=\frac{0.7215 M\times 0.060 L}{0.500 L}[/tex]
[tex]M_2=0.08658 M[/tex]
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 :
[tex]\frac{0.08658}{1000}\times 100 mol=0.008658 mol[/tex] of glucose
Mass of 0.008658 moles of glucose:
[tex]0.008658 mol\times 180.156 g/mol=1.5597 g[/tex]
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Answer:
m_{C_6H_{12}O_6}=1.56g
Explanation:
Hello,
After the dilution, the concentration of glucose is:
[tex]c=\frac{13.0g}{100mL}=0.13g/mL[/tex]
Now, we apply the dilution equation to compute the concentration after the dilution to 0.500L (500mL):
[tex]c_1*V_1=c_2*V_2\\c_2=\frac{c_1*V_1}{V_2}=\frac{0.13g/mL*60.0mL}{500mL} \\c_2=0.0156g/mL[/tex]
Finally, we compute the present grams in 100mL of the 0.0156g/mL glucose solution as shown below:
[tex]m_{C_6H_{12}O_6} =100mL*0.0156g/mL\\m_{C_6H_{12}O_6}=1.56g[/tex]
Best regards.
