Question

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

Answer 1

Hint: This is the numerical problem which needs to be solved step wise. Before that we must know that ppm is the concentration which is defined as the one-part solute per one million parts of solvent.

Complete step-by-step answer: Let us solve this problem step by step;
Given that,
Given mass of strontium chloride hexahydrate = 1 g
Molar mass of strontium chloride hexahydrate = 266.618 g/mol
Molar mass of strontium chloride = 158.526 g/mol
Molar mass of strontium = 87.62 g/mol
We need to find the percent composition of strontium chloride in sodium chloride hexahydrate, using molar masses;
$\dfrac{158.526g/mol}{266.618g/mol}\times 100=59.46%SrC{{l}_{2}}$
Thus, 1 g of strontium chloride hexahydrate contains,
$1g hydrate\times \dfrac{59.46gSrC{{l}_{2}}}{100ghydrate}=0.5946gSrC{{l}_{2}}$
Now, we need to find the percent composition of strontium in strontium chloride, using molar masses;
$\dfrac{87.62g/mol}{158.526g/mol}\times 100=55.27%Sr$
Thus, 0.5946 g strontium chloride contains,
$0.5946gSrC{{l}_{2}}\times \dfrac{55.27gSr}{100gSrC{{l}_{2}}}=0.3286gSr$
Now, to get the concentration of strontium in ppm, we need to divide the mass of solute by the mass of solvent and multiply the ratio by ${{10}^{6}}$. To get the mass of solvent, consider the density of water as 1000 g/L;
Mass of solvent $=100L\times \dfrac{1000g}{L}={{10}^{5}}g$
Thus,
Concentration of strontium ions in ppm $=\dfrac{0.3286g}{{{10}^{5}}g}\times {{10}^{6}}=3.3ppm$

Note: Do note to use proper units and solve with mere presence of mind as this type of problem is easy to understand but much difficult to solve as it can confuse in many ways.
Also, as mentioned earlier in the problem statement solve this in the concentration unit of ppm; not any other units.