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For an aqueous solution of sucrose (C12H22O11), determine: (a) the molarity of 3.30 L of a solution that contains 225.0 g of sucrose M (b) the volume of this solution that would contain 4.30 mole of sucrose L (c) the number of moles of sucrose in 1.00 L of this solution mol C12H22O11

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Answer:

a) Molarity = 0.199M

b) 21.6L of solution

c) 0.199 moles of sucrose in 1L

Explanation:

a) Molarity is the ratio of moles of solute (sucrose) in volume of solution (3.30L):

Moles sucrose -Molar mass: 342,3g/mol-:

225.0g * (1mol / 342,2965g) = 0.6573 moles

Molarity = 0.6573 moles / 3.30L

Molarity = 0.199M

b) 0.199M means 0.199 moles in 1L. 4.30 moles are in:

4.30 mol * (1L / 0.199mol) = 21.6L of solution

c) By definition, there are 0.199 moles of sucrose in 1L. This is the meaning of 0.199M

samueladesida43
  1. The molarity of 3.30 L of a solution that contains 225.0 g of sucrose is 0.2M
  2. The volume of this solution that would contain 4.30 mole of sucrose is 21.5L
  3. The number of moles of sucrose in 1.00L of this solution is 0.2mol

HOW TO CALCULATE MOLARITY:

  • The molarity of a solution can be calculated by dividing the number of moles of the substance by its volume. That is:

  • Molarity (M) = no. of moles ÷ volume

QUESTION A:

  • Molar mass of C12H22O11 = 342.3 g/mol
  • no. of moles = 225g ÷ 342.3g/mol = 0.66mol

  • Molarity of sucrose = 0.66mol ÷ 3.30L = 0.2M

QUESTION B:

  • Volume = no. of moles ÷ molarity

  • Volume = 4.30 mol ÷ 0.2M = 21.5L

QUESTION C:

  • no. of moles = molarity × volume

  • no. of moles = 0.2M × 1L

  • no. of moles = 0.2mol

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