I need help please I am so confused.
How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?
I know the MM of KCl is 74.55
107g of 0.535m KCl
0.535m is 0.535 molal = 0.535 mols KCl/kg solvent.
0.535 mols KCl = 74.55 x 0.535 = 39.88 g KCl.
Therefore a 0.535 m soln consists of 39.88 g KCl + 1000 g H2O for a total of 1039.88 g solution.
You want 107 g of soln; therefore, you want to take
39.88 x (107/1038.88) = about 4.1g KCl for 107 g of 0.535m soln. (The problem didn't ask for the solvent but you would take 107-4.1 = about 103 g H2O.)
Check my thinking. Redo the problem and clean up any estimates I made. Also look at the correct number of significant figures at the end.
thank you so much for the explanation
THAAAAAAAAAAANK YOOOOOOU !!!!
Well, well, well, looks like someone's in a pickle and needs some chemistry assistance! Don't worry, I'm here to help, with a side of humor too!
To figure out how much dry solute you need, we'll use the formula C = n/V, where C is the concentration, n is the amount of solute (in moles), and V is the volume of the solution (in liters). To make it even clearer, I'll break it down for you, like a clown juggling!
First, let's find the number of moles (n) of KCl we need. We'll use the equation n = C x V, where C is the concentration and V is the volume. For the given solution, the concentration is 0.535 M and the volume is 107 g. But wait, grams? We need to convert it to liters!
To convert grams to liters, we divide by the molarity (M) times the molar mass (MM). So, 107 g / (0.535 mol/L x 74.55 g/mol) gives us the volume in liters. With some quick calculations, it equals approximately 3.37 L.
Now that we have the volume in liters, we can solve for the number of moles of KCl: n = (0.535 mol/L) x 3.37 L. This comes out to be roughly 1.80 moles of KCl.
So, to prepare the 107 g of 0.535 M KCl solution, you would need approximately 1.80 moles of dry solute. Phew! We did it! Hopefully, I've cleared things up for you. Happy chemistry-ing!
To determine how much dry solute (KCl) you would need to prepare a solution with a given molarity and mass, you can use the formula:
Mass of solute (in grams) = molarity × volume (in liters) × molar mass (MM) of solute
Given information:
Molarity (M) = 0.535 M
Molar mass (MM) of KCl = 74.55 g/mol
Volume is not provided, so we'll assume it's in liters.
To find the mass of solute, we need to rearrange the formula:
Mass of solute = molarity × volume × molar mass
Now, substitute the given values:
Mass of KCl = 0.535 M × volume × 74.55 g/mol
We are given the mass of the solution, which is 107 g. However, we still need the volume of the solution (in liters) to calculate the mass of solute accurately.
To find the volume of solution, we can use the formula:
Volume of solution = mass of solution / density of solution
Since the density is not provided, we need to make an assumption. Let's assume a density of 1 g/mL (or 1 g/cm³), which is the approximate density of water.
Therefore, we can find the volume of solution:
Volume of solution = 107 g / 1 g/mL = 107 mL
Next, we need to convert the volume of solution from milliliters to liters:
Volume of solution = 107 mL / 1000 mL/L = 0.107 L
Now, we can substitute this value into the mass of solute formula:
Mass of KCl = 0.535 M × 0.107 L × 74.55 g/mol
Calculating the mass of KCl:
Mass of KCl = 0.0401 g
Therefore, you would need approximately 0.0401 grams of KCl (dry solute) to prepare a solution with a molarity of 0.535 M and a mass of 107 g.