When 2.13 g of a nonelectrolyte solute is dissolved in water to make 175 mL of s

Question

When 2.13 g of a nonelectrolyte solute is dissolved in water to make 175 mL of solution at 21 °C, the solution exerts an osmotic pressure of 973 torr.

Map a Sapling Learning acmillan learning When 2.13 g of a nonelectrolyte solute is dissolved in water to make 175 mL of solution at 21 C, the solution exerts an osmotic pressure of 973 torr. What is the molar concentration of the solution? Number How many moles of solute are in the solution? Number mol What is the molar mass of the solute? Number g/ mol

Explanation / Answer

Given :

Mass of substance dissolved = 2.13 g

Volume of solution = 175 mL ; 0.175 L

T = 21 0C = 273.15 + 21 = 294.15 K

Osmotic pressure = 973 torr

* Molar concentration of solution

Use the following formula to calculate molarity

Osmotic pressure = i x M x RT

Here osmotic pressure is in atm , i (Vant’Hoff Factor) is 1 since the solution is non electrolyte

R is gas constant ( 0.0821 L atm / ( K mol)

T (Temperature) = 294.15 K

Convert the pressure into atm

Pressure in atm = 973 torr x 1 atm / 760 torr = 1.2803 atm

Use given value to get molarity

M = 1.2803 atm / ( 0.0821 L atm / ( K mol) ) x 294.15 K) =0.0530 M

* Moles of solute

Moles of solute = volume in L x molarity = 0.175 L x 0.0530 M = 0.009277

= 0.01 mol

* Molar Mass

Number of moles = mass in g / molar mass

Therefore

Molar mass = mass in g / moles = 2.13 g / 0.01 mol = 229.6 g /mol

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