Akron Hungarian club to have 101st grape dance

The Akron Hungarian American Club will celebrate its 101st grape dance from 5 to 10 p.m. Saturday, Oct. 12, at 694 E. Waterloo Road.

The event, open to the public, will include food, music and dancing. Admission fee is $3. The first dance will be at 6 p.m.

A cash-only menu will include sausage sandwiches, sauerkraut, cabbage noodles, Hungarian flatbread and hot dogs.

The club has held a grape celebration every October for more than a century. Members hang grape decorations from the ceiling, don traditional costumes and perform folk dances.

For more information, visit https://akronhungarianamericanclub.com/ or call 330-724-0259.

More: Local history: Hungarian club turns 100 in Akron

This article originally appeared on Akron Beacon Journal: Akron Hungarian club to have 101st grape dance