11 bits resolution (an unusual number, by the way) is equal to 2^11 = 2048 discrete values that a single reading can take. If your card has a range of +/- 10 V, then that's a 20-volt spread, divided into 2048 steps (I'll assume -1024 to +1023). That means you have a resolution of 0.009766 Volts. (20 V / 2048 steps)
If you could set your card to be UNIPOLAR, it would be better, since your signal is unipolar (not negative voltage). By using a 0 to +5V signal into a -10 to +10 Volt range, you are using only one fourth of the range (wasting 2 bits). If the range is -1024 to +1023 counts, your signal is only going to range from 0 to +511. So if you had a reading of 256, for example, the voltage would be 256 * 0.009766 = 2.5000 Volts. Hope that helps.
