Share
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM solution and __
Question
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM solution and ___________ compared to a cell with 300 mOsM (non-penetrating solutes) interior.
in progress
0
Chemistry
3 years
2021-07-15T23:39:06+00:00
2021-07-15T23:39:06+00:00 1 Answers
0 views
0
Answers ( )
Answer:
A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.
Explanation:
The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.
When they are dissolved in water, they dissociate into particles as follows:
NaCl → Na⁺ + Cl⁻ (2 particles per compound)
CaCl₂ → Ca²⁺ + 2 Cl⁻ (3 particles per compound)
urea: not dissociation (1 particle per compound)
Then, we have to calculate the osmolarity of the solution. We multiply the molarity of each compound by the number of particles produced by the compound in water:
Osm = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) + (20 mM urea x 1) = 280 mOsm
Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.
To compare with a cell’s osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl and CaCl₂) is calculated as:
Osm (non-penetrating solutes) = (100 mM NaCl x 2) + (20 mM CaCl₂ x 3) = 260 mOsm
Therefore, we have:
Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution
Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic