Which solution has a higher vapor pressure, 1 M NaCl or 1 M glucose(C6H12O6)?
To determine which solution has a higher vapor pressure, we need to understand the concept of colligative properties. Vapor pressure, in this case, is a colligative property which depends on the number of solute particles present in a given solution.
According to Raoult's Law, the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. In other words, the more solute particles present, the lower the vapor pressure of the solution.
Now, let's compare the number of solute particles in each solution:
1 M NaCl:
NaCl dissociates into Na+ and Cl- ions in water. So, for every 1 mole of NaCl, we have 2 moles of solute particles (Na+ and Cl-). Hence, the total number of solute particles in 1 M NaCl is 2 moles.
1 M glucose (C6H12O6):
Glucose does not dissociate into ions in water; it remains as individual molecules. Therefore, the number of solute particles in 1 M glucose is 1 mole.
Based on the above comparison, we can conclude that 1 M NaCl has a higher number of solute particles compared to 1 M glucose. Consequently, 1 M glucose will have a higher vapor pressure because it has fewer solute particles.
I worked the osmotic pressure almost to completion for you. This one I'll get you started and let you finish.
Psoln = Xsolvent*Posolvent
I wonder if you intended to write 1 M or 1 m; i.e., 1 molar or 1 molal.