The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is .

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.