Question

Please answer questions 3-5

3) A solution has 10 mM of glucose and 25 mM of albumin protein. What is the osmolarity lennso ensidmem of eub stom 1sloc bne.enol begisro 0 ortt wobevon 2eluoolom esen eosevsel 10 Y oe berotnoonoo 2es ot o noituloe belsntnson00 nistog to eeqy 016 tod) enlelo10 1elmeo o lennsrdo ens nelog 26 noiautib bateillost bollso al aeso1 ein 4) A solution contains 50 mM NaCl, 20 mM alucose, and 10 mM albumin. What is the n8onoo Sup9 of this solution? olds osmolarity of this solution? stsdne It is straightforward to compare osmolarity between a cell and a solution. A solution with a higher osmolarity is hyperosmotic to the cell. A solution with the same osmolarity compared to a cell is isosmotic. A solution with a lower osmolarity is hyposomotic. 5) Consider the following cell placed in the solution. Then answer the following questions. l A) Calculate the osm olarity of the cell: B) Calculate the osmolarity of the solution: C) The solution is a) hyperosmotic b) isoosmotic c) hyposomotic Cell Solution ma eieomao 00 lleo er on 15 mM NaCi 5 mM Urea to the cell. stat ent nor tegsvn arl 15 mM Dextrose

Answer 1

Question 3):

The solution has two osmotically active species, and none is a salt, so both moles of both species are going to be the same amount but in osmoles.

So, the amount of osmoles per every 1000 ml ( or 1 liter) of solution it will be:

10x10^-3osmoles due to glucose + 25x10^-3osmoles due to albumin protein

amount of osmoles es equal to 35x10^-3osmoles

Then, the osmolarity will be the amount of osmoles divided by the liters of solution, which is one liter, as stated above.

Osmolarity= 0,035