Can someone explain this to me!?

Question

Suppose there is a computer with an 18 bits address space and the cell size is 8 bits.
What is the smallest and highest address?
What is the largest possible memory size in kilobytes?

I have assumed farmula 2^18 *8 = 2097152bytes which is 2097kb am i doing it right or wrong?
It is not my assignment it is previous semester paper nd i want to solve it to understand nd be prepare for my upcoming exam
Thanks

Answer 1

Hope these helps clearly.. https://blogs.glowscotland.org.uk/nl/ColtnessHS-Computing/tag/data-bus/

https://www.ritambhara.in/how-much-byte-addressable-memory-you-can-use-with-12-bits-address-bus/

https://kb.iu.edu/d/ackw