To prepare 2L of a 5M solution, you should put in 4.6575grams of KMnO4. It is important to make sure that you add them in that order. K should be added first, then Mn.
100mm
To make a solution you add a solute to a solvent.
it is reguarly 5m
4 inches = 100mm
5m = 40m = 40/5m = 8
MW 74.5, 1M solution requires 74.5g in 1L, 5M solution requires 74.5x5 = 372.5g in 1 L. You would need 93.1g KCL to make 250mL 5M solution
In order to have a 2M solution is means that you have 2 moles for every 1 liter. Because you only want to make 800mL of this solution you only need (0.8*2) moles of or 1.6 moles of the compound in your solution. This 1.6 moles is coming from your 5M solution. So how much 5M do you need to take out to have 1.6 moles. A 5 molar solution has 5 moles for every liter, since you only need 1.6 moles of it that you need(1.6/5) liters of the solution or 0.32 liters (320 mL). Calculations done. So your going to take 320 mL of the 5M solution and put it into an 800mL volumetric flask and add water up to the line, or to be more exact, add exactly 460 mL of water in order to dilute the solution from 5M to 800mL of 2M solution.
To prepare 2L of a 5M solution, you should put in 4.6575grams of KMnO4. It is important to make sure that you add them in that order. K should be added first, then Mn.
The pH is 4,14.
You have to evaporate (by open boiling) 45 mL of the 75 mL 2M NaCl solution thus reducing the volume to 30 mL 5M NaCl.
We need 158,3 mL HrPO4 5M.
The answer depends on 5m WHAT! Dollars, cents, rupees, yen?
calculate final molarity of the solution if 11ml of 5m solution is made up to 20ml
What volume(L) of 3M KOH solution can be perpared by diluting 0.5 L of %M KOH solution?
The molecular weight of glucose is 180. Let's say that you intend to make 100 ml of 5M solution. Here is the formula that can help you out every time you are dealing with molarity. (X, which is the desired weight divided by the molecular weight) X (1000/100, which is the desired volume) = 5, which is the desired molarity. So, X/180 X 1000/100 = 5; then X = 90. You need 90 g of glucose to make a 5 M of 100 ml solution of glucose. Hope this is clear, Samba.
.5M