VATICAN CITY, JUL 24, 2001 (VIS) - The itinerary of John Paul II's pastoral visit to Kazakhstan and apostolic visit to Armenia on the occasion of its 1,700 years of Christianity, was published today. The trip starts on September 22 and concludes September 27.
The papal plane will depart from Rome's Fiumicino Airport on September 22 at 8:30 a.m., and arrive at Kazakhstan's Astana International Airport at 7:30 p.m. local time (2:30 p.m. Rome time). The Pope will be in Kazakhstan until the morning of September 25.
Kazakhstan has been an independent republic since 1991. Its capital is Astana. Of the 16,320,000 inhabitants, 47 percent are Muslim, 44 percent are Orthodox, and 1.8 percent are Catholic. It has an archbishop and a bishop, 64 parishes, 45 priests (27 diocesan and 18 religious), 20 seminarians and 53 religious. In 1991, a Latin-rite apostolic administration was established and in 1999, a diocesan structure with one diocese and two apostolic administrators.
The second leg of the Pope's trip starts on September 25 at 11:00 a.m. local time when John Paul II will depart Kazakhstan for Zvartnotz International Airport in Yerevan, Armenia, arriving at 1:00 p.m. local time (10:00 a.m. Rome time). He will be in Armenia through September 27.
Armenia is a republic in Asia Minor, which declared its sovereignty in 1991. Its capital is Yerevan. Of the 3,540,000, inhabitants, 147,000, or 4.1 percent, are Catholic. There is an ecclesiastical circumscription for the Armenian Catholics of Eastern Europe, with its seat in Armenia. It has an archbishop, 29 parishes, 17 priests (1 diocesan and 16 religious), 9 seminarians and 14 religious.
The Holy Father is scheduled to depart Armenia on Thursday, September 27 at 7:30 p.m., arriving at Rome's Ciampino Airport at 8:45 p.m. local time, after a flight of four and a quarter hours.
OP;TRIP KAZAKHSTAN; ARMENIA;...;...;VIS;20010724;Word: 250;
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