it will increase the molarity of the acetic acid
It would mean that you have use too much NaOH to titrate your solution, causing your calculated molarity to be on the higher end of the scale.
molarity of 5% NaCl solution would be 1.25M.
Convert the 200 mol of water to kilograms of NaCl
Molarity = moles of solute/volume of solution Find moles NaCl 55 grams NaCl (1mol NaCl/58.44 grams) = 0.941 moles NaCl Molarity = 0.941 moles NaCl/35 Liters = 0.027 Molarity NaCl ( sounds reasonable as 55 grams is not much in 35 Liters of water, which would be about 17.5 2 liter sodas )
Almost exactly 1 M - to be exactly 1.0M would require 58.5 g NaCl
Because the temp was .3º C off in the beginning and in the end of the experiment, the results would be unchanged since it was constant threw the whole experiment.
The number of moles of solute dissolved in 1 L of a solution would be the molarity. As an example, if you had 2 moles of solute in 1 liter the molarity would be 2M.
molarity of 5% NaCl solution would be 1.25M.
If you concentrate a solution, the molarity (moles/liter) will increase.
Molarity is calculated as moles of solute divided by volume of solution in liters. In this case, you have 2 moles of sodium chloride in a 0.5 liter solution. So the molarity would be 2 moles / 0.5 L = 4 M.
would molarity increase, decrease, or stay the same if the room temperature increased by 5 degrees centigrade
Molarity
It most certainly would affect the molarity of oxalic acid. When you change the volume, i.e. dilute, you decrease the molarity. It is the number of moles that doesn't change.
Isn't this the wrong section? >_> Square brackets generally refer to the concentration of whichever element or compound. For example [A] would refer the the concentration of A (generally calculated in moles/Liter or molarity).
molarity
Molarity
The concentration of a solution can be expressed in many ways. One of them is as the molarity of the solution. A solution with molarity equal to one has one mole of the solute dissolved in every liter of the solutions
The question, as worded, is a little ambiguous. Rather, the question you should be asking is “What is the molarity of a 125 ml aqueous solution containing 10.0g of acetone?” Acetone is roughly 58 grams per mole. Therefore, a 125 mil solution with 10 g of acetone would contain roughly 0.17 moles, and the molarity would be roughly 1.4See the Related Questions for more information about how to calculate the molarity of a solution