How would you prepare 165 mL of a 0.0268 M solution of benzoic acid 1C7H6O22 in chloroform 1CHCl32?

Video transcript

Hey, everyone were asked how we would prepare 2. liters of 0.150 molar solution of barium nitrate. If we started off with solid barium nitrate first, let's go ahead and calculate the mass of barium nitrate required for this reaction. Starting off with our 2.50 liters, we can take our polarity of 0.150 molar and we're going to use our dimensional analysis here, taking the molar mass of barium nitrate. We know that we have 261.347 g of barium nitrate, Her one mole of barium nitrate. Now, when we calculate this out and cancel out all our units, we end up with a total of 98.0 g of barium nitrate. So this is the amount of barium nitrate required. In order to prepare the solution, we would need to weigh 98.0 g of barium nitrate, then dissolve it in a small amount of water in a volumetric flask and dilute it two, L. And this is going to be our final answer for this question. So I hope that made sense. And let us know if you have any questions