Final answer:
The solute potential for a 0.1 M solution of NaCl at 20°C is approximately -9.77 bars, calculated using the van 't Hoff equation with the van 't Hoff factor of 2 because NaCl fully dissociates into two types of ions.
Step-by-step explanation:
The solute potential of a 0.1 M NaCl solution at 20°C can be determined by understanding that NaCl dissociates into two ions, Na+ and Cl-. As NaCl is an ionic compound, the total particle concentration in a 0.1 M solution is actually 0.2 M because it dissociates into two particles per unit of NaCl.
The van 't Hoff equation describes the relationship between solute concentration and solute potential (Y), given by the formula: Y= -MiRT.
Here, M stands for the molarity of the solute, i is the van 't Hoff factor, which is 2 for NaCl as it dissociates into two ions, R is the ideal gas constant, and T is the temperature in Kelvin.
In this scenario, the van 't Hoff factor (i) is 2 because NaCl dissociates completely in water to give two ions, Na+ and Cl-.
To calculate the solute potential, we need to convert the temperature to Kelvin by adding 273 to the Celsius temperature, giving us 293 K.
With an R value of 0.0831 liter bar per mole K, and knowing that the molarity (M) is 0.2 M for the dissolved particles, the solute potential equation becomes: Y = -1 x 0.2 M x 2 x 0.0831 L·bar/mol·K x 293 K
. Therefore, the solute potential of the 0.1 M NaCl solution at 20°C is approximately -9.77 bars.