# How much would a dosage of 1g strontium chloride hexahydrate raise the PPM of strontium in a total water volume of 100L?

##### 1 Answer

That much *strontium chloride hexahydrate* in that much water would increase the concentration of strontium ions by

#### Explanation:

PPM concentration is defined as **one part solute** per **one million parts solvent**.

To determine the concentration in PPM, you need to divide the mass of the solute **in grams** by the mass of the solvent **in grams**, and multiply the ratio by

This is exactly what you would do to get the *percent concentration*, the only difference being that you multiply said ratio by

The important thing to remember about *hydrates* is that when you're adding them to water, you're essentially adding a combination of **solute and solvent** to your existing volume of water.

To find the increase in strontium levels, you need to first determine two things

*the percent composition of strontium chloride in strontium chloride hexahydrate**the percent composition of strontium in strontium chloride*

Let's start with the first one. Use the molar masses of the hydrate and of the anhydrous salt to get the percent composition of strontium chloride in the hydrate

#(158.526color(red)(cancel(color(black)("g/mol"))))/(266.618color(red)(cancel(color(black)("g/mol")))) * 100 = "59.46% SrCl"""_2#

This means that **1 g** of strontium chloride hexahydrate would contain

#1color(red)(cancel(color(black)("g hydrate"))) * ("59.46 g SrCl"""_2)/(100color(red)(cancel(color(black)("g hydrate")))) = "0.5946 g SrCl"""_2#

Now use strontium's's molar mass to find out the percent composition of strontium in strontium chloride

#(87.62color(red)(cancel(color(black)("g/mol"))))/(158.526color(red)(cancel(color(black)("g/mol")))) * 100 = "55.27% Sr"#

This means that **0.5946 g** of strontium chloride would contain

#0.5946color(red)(cancel(color(black)("g SrCl"""_2))) * "55.27 g Sr"/(100color(red)(cancel(color(black)("g SrCl"""_2)))) = "0.3286 g Sr"#

Now, to get the concetration of calcium in ppm, you need to divide this mass by the **total mass of the solute**. Since you're adding less than one gram of water to a volume of 100 L of water, you can safely assume that the volume will not change.

The total mass of water will be - assuming a density of

#100color(red)(cancel(color(black)("L"))) * "1000 g"/color(red)(cancel(color(black)("L"))) = 10^5"g water"#

The concentration in ppm for the strontium ions will be

#(0.3286color(red)(cancel(color(black)("g"))))/(10^5color(red)(cancel(color(black)("g")))) * 10^6 = color(green)("3.3 ppm")#