Density and pressure are directly proportional
[tex]\\ \rm\Rrightarrow \rho_1P_2=\rho_2P_1[/tex][tex]\\ \rm\Rrightarrow 0.680(8)=5\rho_2[/tex]
[tex]\\ \rm\Rrightarrow 5.44=5\rho_2[/tex]
[tex]\\ \rm\Rrightarrow \rho_2=1.088g/\ell[/tex]
Answer:
[tex]1.088\ g/L[/tex]
Explanation:
Step 1: Determine the amount dissolved
[tex]Density_{\ 1}*Pressure_{\ 2}=Density_{\ 2}*Presure_{\ 1}[/tex]
[tex]0.680\ g/L*8\ atm=Density_{\ 2}*5\ atm[/tex]
[tex]\frac{5.44\ g*atm/L}{5\ atm}=\frac{Density_{\ 2}\ *\ 5\ atm}{5\ atm}[/tex]
[tex]1.088\ g/L=Density_{\ 2}[/tex]
Answer: [tex]1.088\ g/L[/tex]
