I did not know that you could get a concentration of 75.66 M KCl, but;
Molarity = moles of solute/Liters of solution
75.66 M KCl = moles KCl/1 liter
= 75.66 moles of KCl
75.66 moles KCl (74.55 grams/1 mole KCl)
= 5640 grams KCl
that is about 13 pounds of KCl in 1 liter of solution. This is why I think there is something really wrong with this problem!
Approximately 18.4 grams of baking soda is required to make 50 ml of a saturated solution at room temperature.
1 and a half times
To prepare a 500mM KCl solution, you would need to dissolve 74.55 grams of KCl in enough solvent to make 1 liter of solution.
To prepare a 1 normal solution of NaOH, you would need to dissolve 40.00 grams of NaOH in enough water to make 1 liter of solution. This is because the molecular weight of NaOH is 40 g/mol.
The concentration of a solution is some measurement of how much solute there is in the solution.
The concentration of a solution is some measurement of how much solute there is in the solution.
To determine the concentration of a solution, you would need to separate the solution. You then determine how much of the solution is diluted, and how much is whole.
The time required to administer 200ml of glucose IV solution would depend on the rate of administration, typically measured in ml per hour. For example, if the rate is set at 100ml per hour, then administering 200ml would take 2 hours.
You think probable to a saturated solution.
This solution is superasaturated.
The answer will depend very much on the nature of the equation. The steps required for a one-step equation are very different from the steps required for a partial differential equation. For some equations there are no straightforward analytical methods of solution: only numerical methods.
Why can evaporation be used to find out how much salt is in a solution