Question

17.which sample contains more molecules, 15.0L of steam (H2O (9)) at 123.0°C and 0.93 atm or 10.5g of ice (H2O (s)) at -5°C? Please show your calculation for the number of molecules in each sample for full credit. (9) 18. a 0.123 g sample of a gas occupies 124mL at 120°C and 760 torr. What is the molar mass of the gas? (6) 19. A gas sample has a volume of 4250mL at 15°C and 745mmHg. What is the new temperature of the gas if the sample is transferred to a new container with a volume of 2.50L and 1.20 atm? (9)

Answer 1

(7) Given dolla. T=183+273-396 k V: 15.0L, P = 8.93 atm Cal. PV = nRT - PV - 0.93 atm x 15.02 RT 0.08206 L'atm x 396k K.mal n=0.43 mol i Imol = 6.023x10 23 molecule Number of molecule = nx 6.023x1023 = 0.43x6023X1029 molecule - 2.58941023 molecule. I cause mass of ice (H2O) = 10.5g molar mass of Hao - 18 g/mol moles of ice = 10.5 g 18.0 g/mal - 0.583 mol Number of molecule = nx6.023x1023 = 0.583 X6.083 X102] 3.511 X 1023 molecule from above calculation it is clear that II 10. 5g ice (H₂O) contain more molecule.

o Given date e m= 0.1238 V= 124 m2 = 12481032 T: 120 +273 = 393k P = 760 forr = larm solution since we know that ideal gas equation, for n mal gas PV = nRT since n = m where m= mass of gas ing. M = molar man inhoud putting the value of n in above equation Pr= m RT M = MRT pv - 0.1239 x 0. 08206 talm x393k ko mal 1 atm x 124 x 1631 = 31.98 M M 32.0 d/mal

Given data Vi= 4250 mL = 4.250 L = 2.50 L Ti = 15+273 = 280k Pi = 745 mm Hg 1 mmHg = 5 atm 45 - 8.980 atm 760 P2 = 1.20 atm We know that Tg = P2 V₂ Ti Pivi = 1.20 atm x 2.50 LX 280k 0.980 atm x 4.250L To = 207.44K 2 or | Tg = 207.44-273 | = -65:56 C