What mass of solute is contained in a $675mL$ volume of solution that was $0.025mol*L^-1$ with respect to $"sodium sulfate"$ ?

Question

1723 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-26T20:09:59

We get a mass of under $2.5*g$ of dissolved solute.

$"Concentration"="Moles of solute"/"Volume of solution"$ , and thus, more symbolically, $C=n/V$ .

As you progress in chemistry, you will use this relationship $"ad nauseum"$ .

Given the formula:

$"moles"="concentration"xx"volume".$ And thus..........

$n_(Na_2SO_4)=0.025*mol*cancel(L^-1)xx675*cancel(mL)xx10^-3cancel(L*mL^-1)$

$=0.016875*mol$ , note that dimensionally, the units we got were $"moles"$ .

And for the mass of this quantity, we perform the product, $"mass"xx"molar mass"$ $=$ $0.016875*cancel(mol)xx142.04*g*cancel(mol^-1)$
$=??g$

What mass of solute is contained in a 675*mL675*mL volume of solution that was 0.025*mol*L^-10.025*mol*L^-1 with respect to "sodium sulfate""sodium sulfate"?