Created using AIUnderstanding mass percent is crucial when working with solutions in chemistry. Mass percent is calculated by taking the grams of solute, dividing it by the grams of the total solution, and then multiplying by 100. This formula allows you to express the concentration of a component in a solution. For example, a 2.50% sodium hydroxide solution means there are 2.50 grams of sodium hydroxide in every 100 grams of the solution. This percentage can also serve as a conversion factor, aiding in the transition between different units of concentration, such as molarity. Remember, the mass percent provides a straightforward method to quantify the ratio of solute to solution, which is essential for precise chemical formulation and reactions.
Mass Percent represents grams of solute per grams of solution, multiplied by 100.
For solutions, mass percent represents the grams of solute per grams of solution multiplied by 100. So when we're looking at the mass percent of solutions, just remember that it equals grams of solute divided by grams of solution multiplied by 100.
So keep this formula in mind anytime we're given a question that deals with solutions and mass percent.
Calculate the mass percent of a solution prepared by dissolving 34.1 grams of potassium chloride in 150 grams of water. All right, so we're going to say here that our mass percent equals the grams of solute. Remember, the solute is the one that's smaller in amount. In this case it would be the 34.1 grams of potassium chloride divided by grams of solution.
Remember, your solution is your solute and your solvent added together. So KCl is our solute plus water, which is our solvent and that would be times 100. When you plug this in, you'll get back as your mass percent 18.5%. So this would represent the mass percent of our given solute shape.
Calculate the amount of water (in kilograms) that must be added to 12.0 g of urea, (NH2)2CO, in the preparation of a 18.3 percent by mass solution. The molar mass of urea, (NH2)2CO, is 60.055 g/mol.
Now it's important to remember that when converting to other unit terms such as molarity, the Max percent value itself can represent a conversion factor. So here for example, when we have 2.50% sodium hydroxide solution, it can be represented as 2.50 grams of sodium hydroxide within 100 grams of solution. So this can be written as 2.50g NaOH100 grams of solution.
Now since it's a conversion factor, we can also do the reciprocal where we have 100 grams of solution on top and 2.50 grams of NaOH on the bottom. So keep this in mind. We can basically expand what the mass percent is whenever a value is given to us. That percentage is that number over 100 grams of solution.
Determine the molarity of a sulfuric acid solution. That is 5% sulfuric acid with a density of 0.9918 grams per milliliter. All right, so remember, molarity itself is capital M, so we need to isolate the units of moles over liters in order to isolate molarity. All right, So remember that when we are given a mass percent, it can be used as a conversion factor as well, so 5%. So sulfuric acid here really means 5 grams of sulfuric acid per 100 grams of solution.
And here we have the density of the solution, which is 0.9918 grams of solution per one milliliter of solution. All right, so how do we use this information to help us get to our molarity? All right. So we're going to start out with our mass percent, which is 5.0 grams of sulfuric acid per 100 grams of solution. All right? And we're going to say here, we want to get rid of the grams of sulfuric acid and isolate moles of sulfuric acid. So for everyone mole of sulfuric acid the molar mass of it is 98.086 grams. Grams here cancel out.
Now I have moles of sulfuric acid over grams of solution. Next I need to get rid of grams of solution because remember molarity has liters on the bottom. We do this by using our density. So here our density is 0.9918 grams of solution per one milliliter of solution. So here grams of solution cancel out and now I have milliliters of solution. I'm almost there. All I got to do now is just change these milliliters into liters. So remember that one milli is equal to 10-3 liters and like that we've isolated moles over liters.
So we've just isolated our molarity. So when we plug that in, we're going to get 0.5056 molar. Now since this has two sig figs, we'll just bring this down at two sig figs as well. So this is 51 molar. So that'll be our molarity for this given solution of sulfuric acid.
A solution was prepared by dissolving 51.0 g of KBr in 310 mL of water. Calculate the mass percent of KBr in the solution.
An aqueous LiNO2 solution is made using 90.3 g LiNO2 and diluting it to a total volume of 1.72 L. If the density of the solution is 1.20 g/mL, what is the mass percent of the solution?
Determine the percent sulfuric acid by mass of a 1.37 m aqueous solution of H2SO4.
