Question

Question 16 Hydrochloric acid is prepared by bubbling hydrogen chloride gas through water. What is the molarity of a solution prepared by dissolving 219 L of HCl(g) at 37°C and 8.91 atm in 5.99 L of water?

Question 17 What is the molar mass of a gas if 1.33 grams of the gas in a 0.497 L flask at 425 K has a pressure of 1.59 atm?

Use R = 0.0821 Latm/molK

Question 19 A 60.3 gram piece of lead at 97.3 °C is placed in a calorimeter containing water at 21.9 °C. If the temperature at equilibrium is 27.6°C, what is the mass of the water

Answer 1

Q16:

PV = nRT

8.91atm×219L = n×0.0821atm-L/K.mol × 310K

n = 76.6685mol

[HCl] = number of moles/volume of solution = 76.6685mol/5.99L = 12.799mol/L

= 12.8 M. (Answer)

Q17:

PV = nRT

1.59atm×0.497L = n × 0.0821atm-L/K.mol × 425K

n = 0.02265mol

Number of moles = mass / molar mass

Molar mass = 1.33g/0.02265mol = 58.726g/mol

= 58.7 g/mol. (Answer)

Q19:

Heat lost by metal = heat gained by water

m1s1∆T1 = m2s2∆T2

60.3g × 0.128J/℃.g × (97.3 - 27.6)℃ = m × 4.184J/℃.g × (27.6-21.9)℃

537.925J = m × 23.8488J/g

m = 22.558g

Mass of the water = 22.6 g. (Answer)