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when 3.12 g of a nonelectrolyte solute is dissolved in water to make 215 ml of solution at 27 °c, the solution exerts an osmotic pressure of 913 torr. what is the molar concentration of the solution? concentration: m how many moles of solute are in the solution? moles of solute: mol what is the molar mass of the solute? molar mass:

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0.0105 mol of non-electrolyte solute, with 297.14 g/mol molar mass, is dissolved in water to make a 0.0488 M solution.

Given:

  • mass of solute (m) = 3.12 g
  • volume of solution (V) = 215 mL
  • Temperature (T) = 27 deg C or 300 K
  • Gas-constant (R) = 0.08206 L-atm/mol-K

Osmotic pressure is expressed mathematically as π=M*R*T. The concentration of the solution "M" should be expressed in molarity, mol/L. Through the use of the formula, let us solve for the concentration of the solution. The osmotic pressure must be converted to atm.

Conversion factor: 1 atm has 760 torr.

913 torr * (1 atm/760 torr) = 1.20131 atm

π = 1.20131 atm = M*(0.08206 L-atm/mol-K)*(300 K)

M = 0.0488 M

From the molar concentration of the solution, we can solve for the number of moles of the solute.

M = mol solute / volume solution

mol solute = M * volume solution

mol solute = 0.0488 M * (215 mL) * (1 L/1000 mL)

mol solute = 0.0105 mol

Then, we can solve for the molar mass of the solute.

molar mass solute = mass solute / mol solute

molar mass solute = 3.12 g / 0.0105 mol

molar mass solute = 297.14 g/mol

To learn more about osmotic pressure, please refer to https://brainly.com/question/8195553.

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