When 3.24 g of a nonelectrolyte solute is dissolved in water to make 955 mL of solution at 22 °C, the solution exerts an osmotic pressure of 973 torr. What is the molar concentration of the solution? concentration:

0.0529mol/L

Explanation:

Step 1:

Data obtained from the question. This includes:

mass (m) = 3.24 g

Van 't Hoff factor (i) = 1 (since the solution is a non electrolyte)

Temperature (T) = 22°C

Volume (V) = 955 mL

Osmotic pressure (Π) = 973 torr

Molar concentration (M) =?

Step 2:

Conversion of the units of the variable to appropriate units. This is so vital because it will help us to obtain the desired result in the right unit.

For temperature:

We need to covert the temperature Fom celsius to Kelvin temperature. This is illustrated below:

K = °C + 273

K = 22°C + 273

K = 295K

For osmotic pressure:

We need to convert the osmotic pressure from torr to atm. This is illustrated below:

760torr = 1atm

Therefore, 973 torr = 973/760 = 1.28atm

Step 3:

Determination of the molar concentration. This is illustrated below:

i = 1

T= 295K

Π = 1.28

M =?

Gas constant (R) = 0.082atm.L/Kmol

Π = iMRT

1.28 = 1 x M x 0.082 x 295

Divide both side by 0.082 x 295

M = 1.28/ (0.082 x 295)

M = 0.0529mol/L

Therefore, the molar concentration of the solution is 0.0529mol/L

The molar concentration of this solution is 0.0529 mol/L

Explanation:

Step 1: Data given

Mass of  a nonelectrolyte solute = 3.24 grams

Volume of solution = 955 mL = 0.955 L

Temperature = 22 °C = 295 K

Osmotic pressure = 973 torr = 973 /760 = 1.28026 atm

Step 2: Calculate the molar concentration

π = i*C*R*T

⇒with π = the osmoti pressure = 1.28026 atm

⇒with i = the van't Hoff factor for a nonelectrolyte = 1

⇒with C = the molar concentration = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 22 °C = 295 K

1.28026 = 1 * C * 0.08206 * 295

C = 1.28026 / (0.08206 * 295)

C = 0.0529 mol / L

The molar concentration of this solution is 0.0529 mol/L

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