Question

6.(14 pts, 8min) Do either (a) or (b) (a) HF(g) has a root mean square speed of 525 m/s. What is the Temperature in "C.of this gas? Data R- 0.0821 L atm/ (K mol)-8.3145 J/(K mol). Atomic masses: H-1.008, F 19.00 (a) If it takes 38.5 s for a given amount of CO(g) to effuse from a chamber, how long, expressed in seconds, will it take the same amount of Ne(g) gas to effuse from this chamber under the same experimental conditions? Atomic masses C 12.01, O-16.00, Ne 20.17 Bonus Problem (4pts, No partial Credit) When 28.88 mL of a concentrated solution of HNO, having a molarity of 2.64 M are added to "z" ml. of water, the resulting solution has a molarity of 0.676 M. What is the value of "z" (expressed in mL)? page - 4

Answer 1

HF gas has a teol. men spare tipeed of 525 mis Expression for vol weon Square velsity of PE Universal ang constant 8,314 7/8 K 1 Teenperature in Kevin M = molecular weight of jaz Metenulan weight of hf 20.01g mot = 20.11770 from 28.01 3x 8.314 525-1 T 20.0110 TE (525)** 20.014102 3x 8.314 T = 221.12 K Ans -51.2°C TOK = -973*€ ] 221.12K - 51.8°C Gas I Molecular weight of colgs = 12:01+ 16 Time taken to effuse = 38.55 Ne Gas 2 : Molecular weight of Ne(a) = 20:17 Graham's Law of Diffusion IR Ratel a L R = rate of diffuusion of gas density of gas 2 Mi= molecular weight of gas 1 My- molecular weight 38.5 , /28.01 of gas 2 20.17 - Ti= time of effusion 38.5x10.17 3 Toe 38 T = time of affusion 179.01 of gas 2 17532.678

28.88 ml Vaume of HNo added. Concentration of HNO - 2.64 M Volume of bond water - z ml Resulting concentration - 0.676M resuiting rolume = (28.88+ 2) mL Then; Using the principle of equi chemical equivalence we have Vp Ni = Vf Nf Viz rolume of solution initially Ni= concentration of solution initially V = roume of solution final Nf = final concentration of solution (28.86 * 2.64) = (22.88+2)x 0.646 (28.78+2) = 360 28,88% 2.64 (28.7877) = - 0.676 - 28.88 ::2.8.88* 2'64 22:0.676 z = 83.905 ml