Có: \(m_{H_2SO^{ }_4\left(1\right)}=\dfrac{200.10}{100}=20\left(g\right)\)
\(m_{H_2SO_4\left(2\right)}=\dfrac{320.15}{100}=48\left(g\right)\)
\(\Rightarrow\Sigma m_{H_2SO_4}=20+48=68\left(g\right)\)
Có: mdd sau= 200+320=520(g)
\(\Rightarrow C\%=\dfrac{68}{520}.100\%\approx13,077\%\)
cốc 1 chứa 200g dd H2SO4 10%
=> C%= \(\dfrac{m_{ct1}}{m_{dd1}}.100\%=10\%\Rightarrow\dfrac{m_{ct1}}{m_{dd1}}=\dfrac{1}{10}\)
\(\Rightarrow m_{ct1}=20g\)
cốc 2 chứa 320g dd h2so4 15%.
=> C%= \(\dfrac{m_{ct2}}{m_{dd2}}.100\%=15\%\Rightarrow\dfrac{m_{ct2}}{m_{dd2}}=\dfrac{3}{20}\)
\(\Rightarrow m_{ct2}=48g\) => mct sau pha trộn là: 20 + 48 = 68 (g) => mdd sau pha trộn là: 320 + 200 = 520 => C% sau pha trộn là: \(\dfrac{68}{520}.100\%\approx13,08\%\)
\(m_{H_2SO_4.10\%}=200\times10\%=20\left(g\right)\)
\(m_{H_2SO_4.15\%}=320\times15\%=48\left(g\right)\)
\(\Rightarrow\Sigma m_{H_2SO_4}=20+48=68\left(g\right)\)
\(\Sigma m_{ddH_2SO_4}=200+320=520\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{68}{520}\times100\%=13,08\%\)
Vậy khi đổ cốc 1 chứa 200g dd H2SO4 10% vào cốc 2 chứa 320g dd H2SO4 15% ta được dd H2SO4 13,08%
