Question

How many molecules of sucrose (C12H22O11, molar mass = 342.30 g/mol) are contained in 14.3 mL of 0.140 M sucrose solution?

Answer 1

Answer: 1.2044*10^21

Explanation:

Molarity ofsucrose solution= 0.140M

Volume of sample= 14.3 mL or 14.3*10^-3 litres

No. Of moles of sucrose = Molarity*Volume

Therefore no. Of moles =0.140*14.3*10^-3= 2*10^-3

1 mole sucrose contains 6.022*10^23 molecules

So 2*10^-3 moles contains 2*10^-3*6.022*10^23= 1.2044*10^21 molecules

Answer 2

We have to find the no of molecules of sucrose are contained in 14.3 ml of 0.14 M sucrose solution.

solution : volume of sucrose solution = 14.3 ml

molarity of sucrose solution = 0.14 M

now, the moles of sucrose in the solution = molarity of solution × volume of solution

= 14.3 × 10⁻³ L × 0.14 M

= 2 × 10⁻³ mol

now no of molecules of sucrose = no of moles of sucrose × Avogadro's number

= 2 × 10⁻³ × 6.023 × 10²³

= 12.046 × 10²⁰

= 1.2046 × 10²¹

therefore the no of molecules of sucrose in the solution is 1.2046 × 10²¹