How memory size are needed if address lines are 9k?

If there are 9,000 address lines, it implies that the system can address (2^{9000}) different memory locations. However, this is an impractically large number since the addressable space would be astronomically high. Instead, if you meant 9 kilobytes (kB), then the memory size would be 9,000 bytes, which is equivalent to 9 kB. For a more precise answer, clarifying the context of "9k" would be helpful.