Fluoride ion is added to drinking water at low concentra- tions to prevent tooth decay. What mass of sodium fluoride (NaF) should be added to 750 L of water to make a solution that is 1.5 ppm in fluoride ion? (LO 13.5) (a) 1.1 g (b) 2.5 g (c) 0.51 g (d) 3.1 g

Video transcript

Hello everyone. So this problem being asked what mass of calcium chloride. So with the formula of C A C L two must be added to 900 liters of water in order to create a solution containing four ppm chloride ion. So let's go ahead and recall that one PPM is the same thing as one mg of our salute Over one L of our solution. So they never just calculated for the milligrams. I saw you. We can go ahead and multiply both sides. By our denominators are the one liter of solution. So we get that. The new equation will be that the milligrams of solu is equal to R P P M multiplied with our solution or the leaders of solution. Alright, so first up here is you calculating the mass of our chloride ion? So we do this in different colors. So we calculate for the mass of our cl minus. So we have our concentration before PPM. So that's just 4.0 mg for every one liter we can convert the we can go ahead and multiply this by the given amount of water. So that's 900 liters. We see that the units of leader will cancel the continuing a dimension analysis. We can convert our milligrams into grounds. So one g for every mg again you can see that the units of milligrams will cancel. So the numerical value that we get is 3.6 units being grams of chloride ions. Now calculate next for the mass of calcium chloride. So the mass of CAC. L two that we're going to go ahead and add. Starting off with the mass of our chloride ion. So 3.6 g of cl minus. We can go ahead and use the atomic mass of chloride Or chlorine. So that's one mole of cl for every 35.45 g of cl. And then we see from our molecular formula for every one mole of our molecules of c a c L two, we need two moles of our chloride ions And then finally we can go ahead and use the molar mass of our molecules are calcium chloride. That is 100 and 10.98 g of C Cell two for every one mole over molecule for units, we can see that the grounds of chlorine ion can cancel the moles council Leaving us with just the units of mass being g. So the numerical value I get. Once I put everything into my calculator, I get the value of 5.6351 units being grams of calcium fluoride. So my final answer then Is that we add 5.64 g of calcium chloride. And this right here is going to be my final answer for this problem