JOHN PAUL II DEPARTS TOMORROW FOR KAZAKHSTAN, ARMENIA

VATICAN CITY, SEP 21, 2001 (VIS) - Pope John Paul leaves Rome's Fiumicino Airport tomorrow morning at 8:30 a.m. for a six-day trip to Kazakhstan and Armenia, the 95th foreign apostolic trip of his pontificate. The papal plane will fly over Italy, Croatia, Hungary, Ukraine, Russia and Kazakhstan, covering 4,200 kilometers in 6 hours. He is scheduled to arrive in Astana, the Kazak capital, at 7:30 p.m., local time (2:30 p.m. Rome time).

Kazakhstan, a former Soviet republic and now part of the Commonwealth of Independent States, declared independence on December 16, 1991. It is the largest of the central-Asian republics, with a territory nine times the size of Italy and a population of only 15 million. Just over half of Kazaks are Sunni Muslims. An estimated 6 million Kazaks are Orthodox Christians. Protestants represent about 2 percent of the population and Catholics number 180,000. There are 4 ecclesiastical circumscriptions, 37 parishes, 4 bishops, 59 priests, 2 permanent deacons, 69 religious and 12 major seminarians. There are an average of 3,051 Catholics per priest.

Armenia is one-tenth the size of Kazakhstan and has a population of 3.8 million. Over 90 percent of the populace belongs to the Armenian Apostolic Church. Christianity was declared in Armenia in 301 and Pope John Paul's visit is to commemorate the 17th centenary of the arrival of Christianity in this nation. Armenia was, in fact, the first state to embrace Christianity as a state religion. Catholics number 150,000. There is one ecclesiastical circumscription. There are 2 bishops, 4 priests, 14 religious, 5 major seminarians and 18 parishes. There are an average of 37,500 Catholics per priest.

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