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

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
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.
An electron i hope this helps
-Ionic bonds lose or gain electrons usually between a non-metal and a metal,
-Covalent bonds share electrons usually between two non-metals
Answer:
Explanation:
Lowering of vapour pressure of solvent is proportional to number of moles of solute dissolved in it per litre .
No of moles of glucose dissolved
= mass dissolved / molecular weight of glucose
= 10 / 180
= .055 moles.
No of moles of sucrose dissolved = 10 / 342
= .029 moles.
So reduction in vapour pressure will be lower of solution dissolving sucrose . It is so because , no of moles of solute dissolved in it is low.