#48*L# of #HCl(g)# under standard conditions of #1*atm#, and #298*K# were dissolved in a #5*dm^3# volume of water. What is #[HCl]# in these circumstances?

1 Answer
May 7, 2017

Answer:

We finally get #[HCl]~=0.4*mol*dm^-3#

Explanation:

From the Ideal Gas equation we can access the number of equiv #HCl(aq)#:

#n=(PV)/(RT)=(1*atmxx48*L)/(0.0821*L*atm*K^-1*mol^-1xx298*K)#

(For your information quotient, #1*dm^3-=1xx(10^-1*m)^3=10^-3*m^3=1/1000*m^3-=1*L#).

We get #n=1.96*mol#

And thus #"concentration"="moles of solute"/"volume of solution"#

#(1.96*mol)/(5*dm^3)=??*mol*dm^3#.