What is the osmolarity of a solution that contains 4.00% (m/v) NaCl (fm = 58.44 g/mole) and 3.00% (m/v) glucose (fm = 180.18 g/mole)?
a) 8.80 osmol
b) 11.0 osmol
c) 1.54 osmol
d) 0.851 osmol
e) 1.37 osmol
f) 7.00 osmol
g) 3.08 osmol
it's neither B or D, i added both osmol's.
.08 osmol/liter NaCl + .03 osmol/liter glucose = 11.00 osmol
not sure what i did wrong
- Bobby_ThinLv 710 months ago
The concentration of NaCl given in the problem is 0.04gmL=40gL
We can divide by the molar mass, getting 40gL⋅mol/ 58.44g≈0.6845 M
The concentration of glucose given in the problem is 0.03gmL=30g.
We can divide by the molar mass, getting 30gL/ 180 g/ mol.≈0.1665 M.
• ϕ is the osmotic coefficient, which accounts for the degree of non-ideality of the solution.
• n is the number of particles (e.g. ions) into which a molecule dissociates.
• C is the molar concentration of the solute;
• the index i represents the identity of a particular solute.
If we ignore ϕ and assume that that everything dissociates perfectly.
This is reasonable because glucose and NaCl generally dissolve near-completely in water.
NaCl dissociates into two ions: Na++ and Cl−−, so nNaCl=2.nNaCl=2.
Glucose, however, does not dissociate, but rather stays as a single molecule.
We now have osmolarity=2*0.6845+1*0.1665=1.5355 osmolar