Question

Plzz help me to find the concentration

Answer 1

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 $\rm H_2SO_4$:

$\begin{aligned}&M({\rm H_2SO_4})\\ =&\; 2\times 1.008 + 32.06 + 4\times 15.999 \\=& \;\rm 98.072\; g\cdot mol^{-1}\end{aligned}$.

Number of moles of $\rm H_2SO_4$ in $\rm 3.92\; g$ of this substance:

$\displaystyle n = \frac{m}{M} = \frac{3.92}{98.072} = \rm 0.0399706\; mol$.

Convert cubic centimeters to liters:

$\rm 200\;cm^{3} = 0.200\; L$.

$\rm 50\;cm^{3} = 0.050\; L$.

$\rm 400\;cm^{3} = 0.400\; L$.

Concentration of this solution:

$\displaystyle c = \frac{n}{V} = \rm \frac{0.0399706\; mol}{0.200\; L} \approx 0.200\; mol\cdot L^{-1}$.

While both the volume and the concentration of the $\rm H_2SO_4$ solution changes when it is diluted, the number of moles of $\rm H_2SO_4$ in this solution will stay the same. Number of moles of $\rm H_2SO_4$ in this $\rm 50\;cm^{3} = 0.050\; L$ of concentrated solution:

$\begin{aligned}n &= c\cdot V\\ &=\rm 0.200\; mol\cdot L^{-1}\times 0.050\; L\\&= \rm 0.01\; mol\end{aligned}$.

That will be the same as the number of moles of $\rm H_2SO_4$ in the diluted solution.

Concentration of this solution after it is diluted to $\rm 400\;cm^{3} = 0.400\; L$:

$\displaystyle c = \frac{n}{V} = \rm \frac{0.01\; mol}{0.400\; L} \approx 0.0250\; mol\cdot L^{-1}$.