A chemist wants to make 5.5 L of a 0.300 M CaCl₂ solution. What mass of CaCl₂ (in g) should the chemist use?

Video transcript

everyone to hear what? Ask. Chocolate and Masterman using iodide That has made it to prepare 0.2 l A 4.59 Mueller of magnesium iodide solution. Recall that polarity. It was moss of solute. But by leaders of solution. Since we're giving the volume and the clarity you can use the maturity equation to solve for moles We have 4.59 Mueller. There's going to be equal to X Divided by 0. leaders. So if you must have 4.59 my word Times 0.2 leaders. Is that equal to X? I'm gonna get X. Is equal to 0.918 moles. And now since we have moles we can convert from moles of magnesium iodide to find grams of magnesium iodide 0. 18 bob a magnesium iodine. And in one mall mechanism I died we had the molar mass which we need to find. The molar mass is equal to the math of Magnesium which is .305g plus two Because there's two iodine comes a mass of iodine Which is .904 gramps. And this will give us 278 113 grams of magnesium. I died. So now this will give us 255 grams. Thanks for watching my video and I hope it was helpful