To prepare 250 mL of a D10/NS solution using D10W and a vial of NaCl 4 mEq/mL, you would need 9.6 mL of NaCl. Normal saline (NS) contains 154 mEq/L of sodium chloride (NaCl). To prepare 250 mL of NS, you would need 250 mL * (154 mEq/L) = 38500 mEq of NaCl. Since the vial of NaCl contains 4 mEq/mL, you would need 38500 mEq / (4 mEq/mL) = 9625 mL = 9.6 mL of NaCl. So the correct answer is b. 9.6ml.