a 0.0260 solution (assume the volumes are additive at these low concentration)
______ L
Please help me. i have no idea where to start
______ L
Please help me. i have no idea where to start

Let x be the necessary volume of water in liters.
(0.310) (0.0440) = (x + 0.310) (0.0260)
Solve for x algebraically:
x = 0.215 L
(0.310) (0.0440) = (x + 0.310) (0.0260)
Solve for x algebraically:
x = 0.215 L