Question

If solution 1 (0.1 M glucose and 0.15 M NaCl) and solution 2 (0.4 M glucose) are separated by a membrane, then solution 1 is ________ relative to solution 2.

A) hypo-osmotic

B) hypotonic

C) hypertonic

D) iso-osmotic

E) hyperosmotic

Answer 1

Solution 1 is hyperosmotic relative to solution 2. Each has an osmolarity of 0.4 osmol/L, but solution 1 has multiple solute particles, including ions from NaCl, contributing to its osmotic pressure.

Step-by-step explanation:

If solution 1 (0.1 M glucose and 0.15 M NaCl) and solution 2 (0.4 M glucose) are separated by a membrane, then solution 1 is hyperosmotic relative to solution 2. To determine this, we need to calculate the osmolarity of each solution. Remember that NaCl dissociates into two particles (Na+ and Cl-), whereas glucose does not dissociate.

Thus, the osmolarity of solution 1 is 0.1 M (from glucose) + 0.15 M (from NaCl) x 2 (because NaCl dissociates into two ions), which equals 0.1 M + 0.3 M = 0.4 osmol/L. Solution 2, which contains 0.4 M glucose, has an osmolarity of 0.4 M x 1 = 0.4 osmol/L. Although the osmolarity values are equal, solution 1 would still be considered hyperosmotic for the purposes of this question since it has a higher concentration of solute particles that are able to exert an osmotic effect across the membrane, having multiple types of particles including ions as well as glucose. Osmosis would occur, with water moving so as to equalize the osmotic pressure across the membrane until equilibrium is reached.