Describe how to prepare each solution from the dry solute and the solvent. c. 125 g of 1.0% NaNO3 solution by mass

Video transcript

Hey everyone. So in this problem we are given a 178 g solution That is 2.3% sodium sulfate by mass. Were asked how is it going to be prepared from dry solute and solvent. So essentially we need to know what the mass value is going to be of both the solvent and the salute in order to form this particular solution. So right off the bat, the easiest way to do this would be by finding out the mass of sodium sulfate that is present in the solution. So it is a 178 g solution. Were given 1 78 gramps since The problem tells us that is, it is a 2.3% sodium sulfate by mass solution. This means that we have 2.3 grams of sodium sulfate for every 100 g of solution of the sample. And this is just based off the percent value because 2.3% is nothing but 2.3 divided by 100 but our grams will cancel out And we will be left with 4. grams of sodium sulfate that we have in our solution. Now, the next step to find out how much of solvent that we need because this is only 4.094 g of the total 178 g that we have in our solution. Well, it's very simple. We just subtract the values 178g -4.094 g of sodium sulfate. Well, not us 1 .906 grams of water. And that will be our solvent in this case. So all this means that we need to dissolve 4.904g for 4.094, sorry, 4.794 g of sodium sulfate Selfie in 173.906 g of water. In order to prepare the solution. I hope this helps. And I look forward to seeing you in the next one.