Answer:
Concentration of the first solution: approximately 0.200 mol/L.
Concentration of the diluted solution: approximately 0.0250 mol/L.
Explanation:
Refer to a modern periodic table for relative atomic mass data:
- H: 1.008;
- S: 32.06;
- O: 15.999.
Formula mass of sulfuric acid [tex]\rm H_2SO_4[/tex]:
[tex]\begin{aligned}&M({\rm H_2SO_4})\\ =&\; 2\times 1.008 + 32.06 + 4\times 15.999 \\=& \;\rm 98.072\; g\cdot mol^{-1}\end{aligned}[/tex].
Number of moles of [tex]\rm H_2SO_4[/tex] in [tex]\rm 3.92\; g[/tex] of this substance:
[tex]\displaystyle n = \frac{m}{M} = \frac{3.92}{98.072} = \rm 0.0399706\; mol[/tex].
Convert cubic centimeters to liters:
[tex]\rm 200\;cm^{3} = 0.200\; L[/tex].
[tex]\rm 50\;cm^{3} = 0.050\; L[/tex].
[tex]\rm 400\;cm^{3} = 0.400\; L[/tex].
Concentration of this solution:
[tex]\displaystyle c = \frac{n}{V} = \rm \frac{0.0399706\; mol}{0.200\; L} \approx 0.200\; mol\cdot L^{-1}[/tex].
While both the volume and the concentration of the [tex]\rm H_2SO_4[/tex] solution changes when it is diluted, the number of moles of [tex]\rm H_2SO_4[/tex] in this solution will stay the same. Number of moles of [tex]\rm H_2SO_4[/tex] in this [tex]\rm 50\;cm^{3} = 0.050\; L[/tex] of concentrated solution:
[tex]\begin{aligned}n &= c\cdot V\\ &=\rm 0.200\; mol\cdot L^{-1}\times 0.050\; L\\&= \rm 0.01\; mol\end{aligned}[/tex].
That will be the same as the number of moles of [tex]\rm H_2SO_4[/tex] in the diluted solution.
Concentration of this solution after it is diluted to [tex]\rm 400\;cm^{3} = 0.400\; L[/tex]:
[tex]\displaystyle c = \frac{n}{V} = \rm \frac{0.01\; mol}{0.400\; L} \approx 0.0250\; mol\cdot L^{-1}[/tex].
