Which contains more solute particles a 0.10 M aqueous solution of methanol (CH3OH) or a 0.10 M aqueous solution of table salt (NaCl)?
Are we supposed to go M*L=moles*(6.022*10^23 solute particles)
No, your supposed to recognize that methanol does not ionize so there is 1 particle of methanol for every particle of the solution. Table salt is NaCl and it ionizes in aqueous solution to form Na^+ and Cl^- so for every molecule of NaCl you will get two particles.
Ah, the age-old battle of methanol versus table salt! Well, my friend, in this case, a 0.10 M aqueous solution of table salt (NaCl) contains more solute particles.
You see, when it comes to solute particles, it's all about the presence of ions. In a solution of table salt, each NaCl molecule breaks apart into one sodium ion (Na+) and one chloride ion (Cl-). So, for every NaCl molecule, you get two solute particles. Double the fun, right?
On the other hand, methanol (CH3OH) does not dissociate into ions in water like NaCl does. So, in a 0.10 M aqueous solution of methanol, you only have one solute particle per molecule of methanol.
So, the table salt solution takes the crown in terms of the number of solute particles! Isn't it remarkable how a simple compound like NaCl can be such a party animal?
Yes, you are correct. The equation M * L = moles * (6.022 * 10^23) can be used to determine the number of solute particles in a given solution.
To compare the number of solute particles in a 0.10 M aqueous solution of methanol (CH3OH) and a 0.10 M aqueous solution of table salt (NaCl), we need to calculate the number of moles in each solution.
Let's assume that the volume (L) of both solutions is the same:
For the 0.10 M aqueous solution of methanol:
Methanol, CH3OH, has a molar mass of approximately 32 g/mol. Since the molarity (M) is 0.10 moles per liter, the number of moles present in 1 liter of the solution is:
moles = M * L = 0.10 mol/L * 1 L = 0.10 mol
Using the Avogadro's number (6.022 * 10^23), we can now calculate the number of solute particles:
number of solute particles = moles * (6.022 * 10^23) = 0.10 mol * (6.022 * 10^23) = 6.022 * 10^22 solute particles
For the 0.10 M aqueous solution of table salt (NaCl):
Table salt, NaCl, has a molar mass of approximately 58.5 g/mol. Similarly, the number of moles present in 1 liter of solution is:
moles = M * L = 0.10 mol/L * 1 L = 0.10 mol
Calculating the number of solute particles:
number of solute particles = moles * (6.022 * 10^23) = 0.10 mol * (6.022 * 10^23) = 6.022 * 10^22 solute particles
Therefore, both solutions have the same number of solute particles, which is 6.022 * 10^22 solute particles.
To determine which solution contains more solute particles, you can use the equation M * L = moles * (6.022 * 10^23) to convert the molarity of the solutions to the number of solute particles.
First, let's calculate the number of solute particles in the 0.10 M aqueous solution of methanol (CH3OH):
Given:
Molarity (M) = 0.10 M
Volume (L) = 1 L (assuming a 1 L solution for simplicity)
Using the formula M * L = moles * (6.022 * 10^23), we can rearrange it to calculate the number of moles:
moles = (M * L) / (6.022 * 10^23)
moles = (0.10 M * 1 L) / (6.022 * 10^23)
moles = 0.10 / (6.022 * 10^23) mol
Now, let's calculate the number of solute particles in the 0.10 M aqueous solution of table salt (NaCl):
Given:
Molarity (M) = 0.10 M
Volume (L) = 1 L (assuming a 1 L solution for simplicity)
Using the formula M * L = moles * (6.022 * 10^23), we can rearrange it to calculate the number of moles:
moles = (M * L) / (6.022 * 10^23)
moles = (0.10 M * 1 L) / (6.022 * 10^23)
moles = 0.10 / (6.022 * 10^23) mol
Comparing the number of solute particles in both solutions, we see that both solutions have the same number of solute particles. This means that a 0.10 M aqueous solution of methanol and a 0.10 M aqueous solution of table salt contain an equal number of solute particles.