Bit capacity calculation from organization A memory has 2048 addressable locations, each storing 8 bits. What is the total capacity of this memory in bits?

Introduction / Context:Translating between addresses, data width, and total capacity is a core skill when sizing memories in digital systems.

Given Data / Assumptions:

• Total addresses = 2048 (which equals 2^11).
• Data per address = 8 bits.
• We are asked for total capacity in bits (not bytes).

Concept / Approach:Total capacity (in bits) = number of addresses * bits per address. No conversion to bytes is needed since the answer is requested in bits.

Step-by-Step Solution:Compute: 2048 * 8 = 16384 bits.Note that 16384 bits = 2^14 bits.Therefore, the correct capacity is 16384 bits.

Verification / Alternative check:As a cross-check, convert to bytes: 16384 bits / 8 = 2048 bytes = 2 KB, which is consistent with 2048 addresses of 1 byte each.

Why Other Options Are Wrong:4096 and 8192: too small; correspond to 2048 × 2 or 2048 × 4, not ×8.32768 and 65536: too large; represent 4 KB and 8 KB in bits, not the specified memory.

Common Pitfalls:Mixing bits and bytes or using decimal “K” instead of binary powers of two.

Final Answer:16384