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When 4.58 g of a nonelectrolyte solute is dissolved in water to make 725 mL of solution at 25 °C, the solution exerts an osmotic pressure of 875 torr.

Required:
a. What is the molar concentration of the solution?
b. How many moles of solute are in the solution?
c. What is the molar mass of the solute?

Respuesta :

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Answer:

a. 0.047 M

b. 0.6778 moles

c. 6.76 g/mol

Explanation:

The formula for osmotic pressure is:

  • π = i*M*R*T

Where:

  • i is the van't Hoff's factor (in this case is 1, because it is a non electrolyte solute)
  • M is the molar concentration
  • R is the ideal gas constant
  • T is the temperature in Kelvin

a.) So we solve for M:

  • 875 torr ⇒ 785/760 = 1.15 atm
  • 25 °C ⇒ 25+273.16 = 298.16 K
  • 1.15 atm = 1 * M * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
  • M = 0.047 M

b.) We calculate the moles of solute, using the volume and molar concentration:

  • 725 mL / 1000 = 0.725 L
  • 0.047 M * 0.725 L = 0.6778 moles

c.) We calculate the molar mass, using the mass and the number of moles:

  • 4.58 g / 0.6778 moles = 6.76 g/mol
Eduard22sly

A. The molar concentration of the solution is 0.047 M

B. The mole of the solute in the solution is 0.034 mole

C. The molar mass of the solute is 134.7 g/mol

A. Determination of the molar concentration of the solution.

Osmotic pressure (π) = 875 torr = 875 / 760 = 1.15 atm

Temperature (T) = 25 °C = 25 + 273 = 298 K

Van't Hoff's factor (i) = 1 (non-electrolyte)

Gas constant (R) = 0.0821 atm.L/Kmol

Molarity (M) =?

π = iMRT

1.15 = 1 × M × 0.0821 × 298

1.15 = M × 24.4658

Divide both side by 24.4658

M = 1.15 / 24.4658

M = 0.047 M

B. Determination of the mole of the solute.

Molarity = 0.047 M

Volume = 725 mL = 725 / 1000 = 0.725 L

Mole of solute =?

Mole = Molarity x Volume

Mole of solute = 0.047 × 0.725

Mole of solute = 0.034 mole

C. Determination of the molar mass of the solute

Mole of solute = 0.034 mole

Mass of solute = 4.58 g

Molar mass of solute =?

Molar mass = mass / mole

Molar mass of solute = 4.58 / 0.034

Molar mass of solute = 134.7 g/mol

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