# 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?

May 7, 2017

We finally get $\left[H C l\right] \cong 0.4 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$

#### Explanation:

From the Ideal Gas equation we can access the number of equiv $H C l \left(a q\right)$:

$n = \frac{P V}{R T} = \frac{1 \cdot a t m \times 48 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 298 \cdot K}$

(For your information quotient, $1 \cdot {\mathrm{dm}}^{3} \equiv 1 \times {\left({10}^{-} 1 \cdot m\right)}^{3} = {10}^{-} 3 \cdot {m}^{3} = \frac{1}{1000} \cdot {m}^{3} \equiv 1 \cdot L$).

We get $n = 1.96 \cdot m o l$

And thus $\text{concentration"="moles of solute"/"volume of solution}$

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