Question

How many grams of sucrose (C12H22O11) are in 1.55 L of 0.728 M sucrose solution?

Answer 1

First calculate the number of moles of sucrose (C12H22O11) as follows:

0.728 M = 0.728 mol / 1.0 L

number of moles of sucrose = 1.55 L *0.728 mol / 1.0 L

= 1.1284 moles of sucrose

Now calculate the amount of 1.1284 moles of sucrose by its molar mass:

molar mass of sucrose= 342.2965 g/mol

the amount of 1.1284 moles of sucrose

= 1.1284 moles of sucrose*342.2965 g/mol

= 386.25 g

Hence there are386.25 g sucrose (C12H22O11) are in 1.55 L of 0.728 M sucrose solution