How many grams of CaCl 2 (molar mass = 111g/mol) are needed to prepare 100. mL of 0.100 M Cl − (aq) ions?

Welcome back, everyone. We need to calculate the mass of bromide ions in a 400 mL solution of 4000.300 molar Cobalt two bromide because we are given the molar of Cobalt two bromide. Recall that polarity is expressed as moles of our substance being dissolved, which is our solute per liter of our solution. So the volume of our solution, we can begin by finding the moles of Cobalt two bromide by multiplying both sides by the volume. So that that term will cancel out on the right. And for the moles of Cobalt two bromide, we would plug in our polarity which is given as 0.300 molar of Cobalt two bromide. And this is multiplied by the volume of our solution given in our prompt as 400 mL. However, we need to convert this to liters by multiplying by our conversion factor where we would recall that one mL in the denominator has an equivalence of 10 to the negative third power of our base unit leaders. So canceling out milliliters, we're left with molar multiplied by leaders. And let's also recognize that we can interpret molar or molar concentration again as moles per leader of solute, meaning that we can now also cancel all our units of liters. So now we are left with moles of cobalt two bromide. And we would find from the products that we have 0. moles of cobalt two bromide. Now, with the moles of our Cobalt two bromide, we're then going to find the mass of our bromide ions by going through conversions. So beginning with our 0.1 12 moles of Cobalt two bromide, we need to then multiply to go from moles of cobalt to bromide to moles of our bromide anion. Now, in order to figure out what this ratio would be, we're going to consider the fact that Cobalt two bromide consists of Cobalt a metal and then bromine which is a non metal. So we have ultimately an ionic compound. And in reference to our solubility rules for bromides, we would recognize this as a soluble bromide. So this is aqueous and being an ionic compound. This is a salt that is going to dissociate into the Cobalt two plus ion as well as our two moles of bromide anions. And from this expression, we can observe that we have a ratio where for one mole of Cobalt two bromide, we have two moles of our bromide ions dissolved. So canceling out our units of moles of Cobalt two bromide, we now have moles of bromide in which we now need to get a mass. So our next multiplier is going to be going from moles of bromide in the denominator to moles, sorry to grams of bromide in the numerator. And this is where we would recall from our periodic table. The molar mass of bromine, which we should recall is equal to 79. g of bromine per mole of bromine. So this allows us to then cancel out our units of moles of bromine leaving us with grams of bromide in our numerator. And in simplifying this expression, we would find the mass of bromine equal to a value of 19.177 g of bromide, which we can round to about three sigs as 19. I'm sorry, that should be as 19.2 g of bromide. So 19.2 g of bromide is our final answer. I hope that this made sense and let us know if you have any questions.

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How many grams of CaCl₂ (molar mass = 111g/mol) are needed to prepare 100. mL of 0.100 M Cl⁻(aq) ions?

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How many grams of CaCl₂ (molar mass = 111g/mol) are needed to prepare 100. mL of 0.100 M Cl⁻(aq) ions?