# How does a 8 bit register bank hold 8 regsiters of 8 bits each?

In 8051, there are 128 bytes of RAM. Of which, 32 are set aside for registers in the form of banks, 8 registers to each bank. Each bank is addressed from say 00 to 07, 08 to 0F, etc, that is, occupying 8 bits each, with each but corresponding to a register, thus 8 registers in each bank.

How can this be possible, when the registers themselves occupy 8 bits? Won't each bank occupy 8*8 = 64 bits instead of 8?