Question 7 chap 4 A 4.00L sample of hydrogen chloride (HCl) gas at 2.40 atm and

Question

Question 7 chap 4
A 4.00L sample of hydrogen chloride (HCl) gas at 2.40 atm and 34°C is completely dissolved in 702 mL of water to form hydrochloric acid solution. Calculate the molarity of the solution. Assume no change in volume. Question 7 chap 4
A 4.00L sample of hydrogen chloride (HCl) gas at 2.40 atm and 34°C is completely dissolved in 702 mL of water to form hydrochloric acid solution. Calculate the molarity of the solution. Assume no change in volume.
A 4.00L sample of hydrogen chloride (HCl) gas at 2.40 atm and 34°C is completely dissolved in 702 mL of water to form hydrochloric acid solution. Calculate the molarity of the solution. Assume no change in volume.
A 4.00L sample of hydrogen chloride (HCl) gas at 2.40 atm and 34°C is completely dissolved in 702 mL of water to form hydrochloric acid solution. Calculate the molarity of the solution. Assume no change in volume.

Explanation / Answer

we have:

P = 2.4 atm

V = 4.0 L

T = 34.0 oC

= (34.0+273) K

= 307 K

find number of moles using:

P * V = n*R*T

2.4 atm * 4 L = n * 0.08206 atm.L/mol.K * 307 K

n = 0.3809 mol

This is number of moles of HCl dissolved in 702 mL water

volume of water = 702 mL = 0.702 L

now use:

Molarity = mol of HCl / volume of water

= 0.3809 mol / 0.702 L

= 0.543 M

Answer: 0.543 M

Feel free to comment below if you have any doubts or if this answer do not work

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.