Chemistry Forum

175.0 g pure \(H_2O\) was placed in a beaker and thereafter 9.80 g pure \(H_2SO_4\) was added and stirred. What is the total volume of the solution and what is the percentage concentration of \(H_2SO_4\) in the solution?

My answer: 1.0 ml= 1.0 g \(H_2O\) Hence 175.0 g pure\(H_2O\)=175.0 ml\(H_2O\)
9.80 g \(H_2SO_4\)=5.36 ml because density of \(H_2SO_4\) is \(1.8302 g/cm^3\)

Hence total volume of the solution= 175 +5.36 =180.36 ml and % conc. of \(H_2SO_4= \frac{5.36}{180.36}=2.97 \)%

Is this answer correct?

No, because percentage of acids are mass percentage. It is V₁ + V₂ not equal sum volume. It is 9,8 g/(175g +9,8g)*100% = 5,3%w.w

To get the volume you need specific gravity of the 5.3% acid.

How to compute specific gravity of 5.3 % w/w \(H_2SO_4 ? \) I know specific gravity =\(\frac{density_{substance}}{density_{water}}\)

Density of water=1.
Density of \(H_2SO_4=1.8302/cm^3\)

How to use specific gravity of 5.3% w/w \(H_2SO_4\) for computing volume of the solution?

This you cannot compute. This you have to look up in a table book. Because volumes of the most compounds are not additative. During mixing you get volume contraction.

https://www.mt.com/dam/mt_ext_files/Edi ... _e_gif.gif