VATICAN CITY, APRIL 19, 2011 (Zenit.org).- The Vatican announced the schedule for Benedict XVI’s June 4-5 apostolic trip to Croatia for the National Day of Croatian Catholic Families.
The Pope is scheduled to depart from Rome’s Fiumicino Airport at 9:30 a.m. on June 4, and arrive at Pleso International Airport in Zagreb, Croatia at 11:00 a.m. where he will take part in a welcoming ceremony.
The Pontiff will make a courtesy visit to Croatian President Ivo Josipović at the presidential palace.
Then, the Holy Father will meet with Prime Minister Jadranka Kosor at the apostolic nunciature at 1:50 p.m.
At 6:15 p.m., in the National Croatian Theater of Zagreb, Benedict XVI will meet with representatives of civil society, the political, academic, cultural, and business worlds, the diplomatic corps, and religious figures.
Later, at 7:30 p.m., the Pope will preside over a prayer vigil with the Croatian youth in Zagreb’s Ban Josip Jelacic Square.
The next day, the National Day of Croatian Catholic Families, the Pontiff will preside over Mass at 10:00 a.m. in the Zagreb Hippodrome.
At 2:00 p.m. he will have lunch with the Croatian prelates at the office of the secretary of the Croatian Bishops’ Council.
The Holy Father will preside over vespers at 5:00 p.m. with the bishops, priests, religious, and seminarians of that country.
At Zagreb’s cathedral dedicated to the Assumption of the Blessed Virgin Mary and to St. Stephen, Benedict XVI will pray at the tomb of Blessed Aloysius Viktor Stepinac, who was the archbishop of Zagreb from 1937 to 1960. The cardinal was persecuted by the communist regime of Josip Tito and died a martyr.
The Pope will then meet with the current archbishop of Zagreb, Cardinal Josip Bozanić, at his residence.
At 7:45 p.m. the Pontiff will leave Pleso Airport for Rome, where he is due to arrive at the Ciampino Airport at 9:15 p.m.
