Given a $6.174g$ mass of sodium sulfate that is dissolved in a $1L$ volume, what is the concentration in terms of $"parts per million"$ with respect to sodium ion?

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Answer 1 · 2017-07-31T12:16:49

With respect to sodium ion, we have a $"2000 ppm"$ concentration.

$"1 ppm"$ specifies $1*mg$ of solute per litre of solution......

We has $((6.174*g)/(142.04*g*mol^-1))/(1*L)=0.0435*mol*L^-1$ with respect to sodium sulfate....

And thus $2xx0.04346*mol*L^-1=8.693xx10^-2*mol*L^-1$ , with respect to sodium ion. Why did I double the concentration of sodium sulfate?

And thus, with respect to sodium ion, we have a concentration by mass of ........

$8.693xx10^-2*mol*L^-1xx22.99*g*mol^-1=2.000*g*L^-1$ .

But $"1 ppm"$ $=$ $1xx10^-3*g*L^-1$ , i.e. $1*mg*L^-1$ , and thus we have.....

$2.000*g*L^-1xx10^3*mg*g^-1=2000*mg*L^-1-=2000$ $"ppm"$ .

Given a 6.174*g6.174*g mass of sodium sulfate that is dissolved in a 1*L1*L volume, what is the concentration in terms of "parts per million""parts per million" with respect to sodium ion?