Decrypt simple CBC message with known IV using XOR encryption
That's easy: $C_1 = K \oplus (M_1 \oplus C_0)$. You know $C_1$, $M_1$ and of course $C_0$, the IV. You can easily rewrite this as $K = C_0 \oplus C_1 \oplus M_1$.

What is left is simply decryption with $K$. In case of XOR encryption, decryption is the same as encryption, so $M_i = (K \oplus C_i) \oplus C_{i-1}$.



Notes:


XOR is associative and commutative, so you can leave out the parentheses in the formulas and rewrite the formulas like I did above.
this will obviously not work with secure block ciphers since retrieving $K$ is not supposed to be possible, even when given a known plaintext / ciphertext pair.