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1998 Kremlin Cup – Women's Doubles

Women's Doubles
1998 Kremlin Cup
1997 Champions Spain Arantxa Sánchez Vicario
Belarus Natasha Zvereva
Champions France Mary Pierce
Belarus Natasha Zvereva
Runners-up United States Lisa Raymond
Australia Rennae Stubbs
Final score 6–3, 6–4
Events
Singles men women
Doubles men women
← 1997 · Kremlin Cup · 1999 →

Arantxa Sánchez-Vicario and Natasha Zvereva were the defending champions but only Zvereva competed that year with Mary Pierce.

Pierce and Zvereva won in the final 6–3, 6–4 against Lisa Raymond and Rennae Stubbs.

Champion seeds are indicated in bold text while text in italics indicates the round in which those seeds were eliminated.


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