Question

A 1.78−L sample of hydrogen chloride (HCl) gas at 2.09 atm and 22°C is completely dissolved in 699 mL of water to form hydrochloric acid solution. Calculate the molarity of the solution. Assume no change in volume.

Answer 1

Answer:

$M=0.960M$

Explanation:

Hello,

In this case, it is firstly necessary to compute the dissolved moles of hydrogen chloride into the water as shown below:

$n_{HCl}=\frac{PV}{RT}=\frac{2.09atm*1.78L}{0.082\frac{atm*L}{mol*K}*295.15K}=0.671molHCl$

Thus, the molarity is computed as shown below:

$M=\frac{0.671mol}{0.699L}=0.960M$

Wherein no change in volume is considered, therefore the volume of the solution was the same volume of water.

Best regards.