Mazeppa (Russian: Мазепа) is a 1909 Russian drama film directed by Vasili Goncharov.[1][2][3][4]

Mazeppa
Russian: Мазепа
Directed byVasili Goncharov
Written by
Produced byAleksandr Khanzhonkov
Starring
CinematographyVladimir Siversen
CountryRussian Empire
LanguageRussian

Plot edit

Mazeppa (1909)

The film tells about the Hetman named Mazepa, who is in love with Kochubey's daughter, Maria, and asks her father for consent to marry her, but his father refuses him. This does not stop them and they run away...[5]

Cast edit

References edit

External links edit