Question

0.2 g molecules of sugar is present in 2 L aqueous solution. Molarity of this
solution is​

Answer 1

Answer:

Given molarity of the solution = 0.2

Volume of solution = 200 ml = 0.2L

Molarity = Moles of solute ÷ Volume in litres

Moles of solute = 0.2 × 0.2 = 0.4

Number of moles = Given mass ÷ Molar mass

Molar mass of C₆H₁₂O₆ = 6 × 12 + 12 + 6 × 16 = 180 g

Mass of glucose in the solution = 180 × 0.04 = 7.2 g

Explanation:

Keep Smiling...

Answer 2

Answer:

=> Given molarity of the solution = 0.2.

=> Volume of the solution = 200 ml = 0.2L.

=> Molarity = Moles of sokute ÷ Volume in litres.

=> Moles of solute = 0.2 × 0.2 = 0.4.

=> Number of moles = Given mass ÷ Molar mass.

=> Molar mass of C6H12O6 = 6 × 12 + 12 + 6 × 16 = 180 g.

=> Mass of glucose in the solution = 180 × 0.04 = 7.2 g.

Explanation:

Keep Smiling...