Question

62) Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you were to do the following? (a) Triple the Kelvin temperature while holding the volume constant (b) Reduce the amount of gas by 1/3 while holding the temperature and volume constant (c) Decrease the volume by 45% at constant T (d) Halve the Kelvin temperature, and triple the volume 63) Assuming that air contains 78.0 percent Ni, 20.0 percent O2, 1.00 percent H2 and 1.00 percent Ar, all by volume how many moles of euch type of gas are present in 1.00 L of air at STP? 64) What is the density of carbon tetrachloride vapor at 714 torr and 125 °C? (the gas constant, R-0.08206 L-atm/mol-K) 65) Pure oxygen gas was first prepared by heating mercury (II) oxide, Hgo: What volume in liters of oxygen at STP is released by heating 10.57 g of HgO? 66) Assume that you are using an open-end manometer filled with mineral oil rather than mercury. What is the gas pressure in the bulb in millimeters of mercury if the level of mineral oil in the arm connected to the bulb is 237 mm higher than the level in the arm connected to the atmospherie pressure is 746 mm Hg? The density of mercury is 13, g/mL, and the density of mineral oil is 0.822 g/mL 746 mm Hg Gas 237 mm mineral oil 67) How many moles of solute are present in each of the following solutions? a) 50.0 mL of 1.20 M HNO b) 40.0 mL of 0.65 M glucose (CaH:0) 0 ml sample of vinegar (dilute acetic acid, CH CO-H) is titrated and found to react with 94.7 mL of 0.200 M NaOH. What is the molarity (M) of the acetic acid solution? The reaction is NaOH(aq)+ CH CO H(aq)CHCO-Na(ag)+H:O0)

Answer 1

Q62)

(a) Pressure of gas is directly proportional to temperature of gas at constant volume . PV = n RT , on increasing temperature to thrice, pressure becomes thrice. P2 = 3× P1.

Pressure becomes 3 times of initial pressure.

(b) at constant T and V. Pressure is directly proportional to number of moles. PV = n RT.

On decreasing number of moles by1/3 of initial value, n2 = (2/3)×n1. pressure also reduces by 1/3 of initial pressure.

P2 = (2/3)×P1.

Pressure decrease to (2/3)times of initial pressure.

(c) V2 = V1 - 0.45×V1 = 0.55×V1., V2/V1 = 0.55

P1V1 = P2V2

P2 = P1×(V1/V2) = P1× (1/0.55) = 1.82×P1

Pressure increase to 1.82 time the initial pressure.

(d) V2 = 3V1, T2 =( T1)/2

P1V1/T1 = P2V2/T2

P2 = P1 × (V1 / V2)(T2/T1) = P1 × (1/3)×(1/2) = (1/6)×P1.

Pressure decrease to 6 times the initial pressure.

Q63) at STP , 1 mole of gas occupies 22.4L.

1 L of air will have (1/22.4) moles.

Number of moles of N₂ = (78.0/100.)(1/22.4) = 0.0348mol

Number of moles of O₂ = (20.0/100.)(1/22.4) = 0.00893mol

Number of moles of H₂ = (1.00/100.)(1/22.4) = 0.000446mol

Number of moles of At = (1.00/100.)(1/22.4)= 0.000446mol

Q64) PM = d RT , where M is molar mass of CCl4, d is density of gas.

(714torr/760torr)atm × 153.823g/mol = d × 0.08206atm-L/K.mol × (273.15 + 125)K

d = 4.423g/L

Density of CCl₄ = 4.42g/L . (answer)

Q65) moles of HgO heated = 10.57g / 216.589g/mol = 0.048802mol.

Moles of O₂ produced = 1/2 × moles of HgO = 1/2 × 0.048802 = 0.024401mol

Volume of O₂ at STP = 0.024401mol × 22.4L/mol = 0.5466L

Volume of O₂ = 0.5466L (answer)