Answer:
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.
