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Russian billionaire pays $88 million for daughter's home

A Russian billionaire recently purchased a home for his daughter in New York City for $88 million.

A Russian billionaire recently purchased a home for his daughter in New York City for $88 million.

When former Citigroup chairman Sandy Weill listed his New York City home for $88 million in November, many thought it would take a while to sell the property, but the 6,744-square-foot apartment was recently purchased by a Russian billionaire for his daughter.

Moving trucks may be seen soon outside of the property, following Dmitriy Rybolovlev purchasing the apartment from Weill, who has vowed to donate the proceeds of the sale to charity. The apartment will house Rybolovlev's daughter, who is studying at an undisclosed university in the United States.

The transaction between Weill and Rybolovlev set a record for the highest individual transaction in New York City history, which was previously set by investor J. Christopher Flowers when he purchased a townhouse for $53 million.

The apartment features 10 rooms, including four bedrooms, a terrace that wraps around more than 2,000 square feet and two wood burning fireplaces.

“This sale is an outlier. It works out to be about $13,000 per square foot, the highest on record, for anything, that has ever occurred,” Jonathan Miller, chief executive of real estate appraiser Miller Samuel, said to Yahoo. “What is ironic is that when Sandy Weill bought it for less than half this amount, he paid the highest price per square foot to date in that building, around $6,400 per square foot. He is again setting a record.”

The Russian billionaire is the 93rd richest man in the world, according to the Huffington Post. With the average price of a home in New York City rounding out at around $455,000, Rybolovlev's new property is at the top of the list in value.