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

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Answer 1 · 2017-05-07T21:31:21

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

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 $"concentration"="moles of solute"/"volume of solution"$

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

48*L48*L of HCl(g)HCl(g) under standard conditions of 1*atm1*atm, and 298*K298*K were dissolved in a 5*dm^35*dm^3 volume of water. What is [HCl][HCl] in these circumstances?