Question

Plzz help me to find the concentration

3.92 g of H2SO4 is dissolved in distilled water of 200cm^3. find the concentration of H2SO4.
if 5ocm^3 of this solution is taken out and distilled watef is added to it until it reaches 400cm^3..find the concentration of H2SO4 in it ​

Answer 1

Concentration of the first solution: approximately 0.200 mol/L.

Concentration of the diluted solution: approximately 0.0250 mol/L.

Step-by-step 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* 1.008 + 32.06 + 4* 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 = (m)/(M) = (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 = (n)/(V) = \rm (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)* 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 = (n)/(V) = \rm (0.01\; mol)/(0.400\; L) \approx 0.0250\; mol\cdot L^(-1)`.