Label 0.15 m cacl2 as isotonic, hypotonic, or hypertonic in comparison to 0.9% nacl (0.15 m nacl).

To know if the solutions are isotonic, hypertonic or hypotonic, we have to determine the osmotic pressure of each solution.

P = iMRT
where
i is the number of ions dissociated for strong electrolytes
M is the molarity
R is the universal gas constant
T is the absolute temperature

For 0.15 M CaCl₂, i = 3 (1 for Ca⁺ and 2 for Cl⁻).
P = (3)(0.15)RT= 0.45RT

For 0.15 NaCl, i = 2 (1 for Na⁺ and 1 for Cl⁻).
P = 2(0.15)RT = 0.3RT

Therefore, the P for CaCl₂ is GREATER than the P for NaCl. Hence, the CaCl₂ solution is hypertonic.