<u>Answer:</u> The mass of glucose in final solution is 0.420 grams
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:
.........(1)
Initial mass of glucose = 10.5 g
Molar mass of glucose = 180.16 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:

To calculate the molarity of the diluted solution, we use the equation:

where,
are the molarity and volume of the concentrated glucose solution
are the molarity and volume of diluted glucose solution
We are given:

Putting values in above equation, we get:

Now, calculating the mass of final glucose solution by using equation 1:
Final molarity of glucose solution = 0.0233 M
Molar mass of glucose = 180.16 g/mol
Volume of solution = 100 mL
Putting values in equation 1, we get:

Hence, the mass of glucose in final solution is 0.420 grams