Question

Calculate the molarity of 0.205 L of a solution that contains 156.5 g of sucrose

Answer 1

2.23 M

Step-by-step explanation:

Molarity is given as moles of solute per liter of solution (
`(mol)/(L)`)

To solve this problem, the molecular weight (M.W.) of sucrose (
`C_1_2H_2_2O_1_1`) must be calculated:

C: 12.01
`(g)/(mol)`

H: 1.01
`(g)/(mol)`

O: 16.00
`(g)/(mol)`

Total M.W.= 12(12.01
`(g)/(mol)`)+22(1.01
`(g)/(mol)`)+11(16.00
`(g)/(mol)`)=342.34
`(g)/(mol)`

Moles of sucrose can be found through dimensional analysis:

`156.5g((1mol)/(342.34g))=0.457mol`

Molarity can be found with:

`Molarity=(0.457mol)/(0.205L)\\Molarity=2.23M`