Given $1.0molL^-1$ solutions of $NaCl(aq)$ , what are the masses of solute in.....?

Question

$(i)$ $"a 1 L volume of solution;"$
$(ii)$ $"a 250 mL volume of solution;"$
$(iii)$ $"a 100 mL volume of solution?"$

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Answer 1 · 2017-09-18T18:30:53

Well the molar mass of $NaCl$ is $(22.99+35.45)*g*mol^-1=$ $58.44*g*mol^-1$ .......

From where did I get the atomic masses?

For the remaining problems, we use the relationship.....

$"Concentration/Molarity"="Moles of solute"/"Volume of solution"$

And so, if we gots a $1*L$ volume of $1*mol*dm^-3$ solution with respect there is a mass of.....

$1*Lxx1*mol*L^-1xx58,44*g*mol^-1=58.44*g$

And note that $1*dm^3-=(1xx10^-1*m)^3-=1xx10^-3*m^3=1*L$ , because there are $1000*L$ in a $m^3$ .

And we also gots $250*cm^3xx10^-3*dm^3*cm^-3*2*mol*dm^-3xx58.44*g*mol^-1=29.22*g$

And I will let you the third.......