The following experiment is carried out in a physiology lab. Red blood cells are placed in different solutions: distilled water, 0.45% saline, 0.9% saline, 5% urea, and 5% glucose. The molecular mass for each molecule is given: NaCl = 58.44 gm/mole; urea = 60.06 gm/mole; glucose = 180.16 gm/mole.
For each of the solutions used, calculate the osmolarity. To do this, you will need to convert solutions given in percents to osmolarity. A 0.45% saline solution has 0.45 gm/100 mL NaCl, or 4.5 gm/L. The same is true for the other molecules.
Fill in the osmolarity of each solution to the nearest integer using the choices provided.
Distilled water = ... mOsm
0.45% saline = ... mOsm
0.9% saline = ... mOsm
2.0% urea = ... mOsm
5% glucose = ... mOsm

Osmolarity of distilled water = 0

Osmolarity of NaCl=154 Osm

Osmolarity of NaCl=0.308 Osm

Osmolarity of urea=1.66 Osm

Osmolarity of urea=0.66 Osm

Osmolarity of glucose=0.556 Osm

Explanation:

osmolarity is calculated by the following expression:

O=2*molarity of salt

Osmolarity of distilled water = 0

Osmolarity of 0.45% NaCl:

moles of NaCl=weight/molecular weight=0.45/58.44=7.7x10^-3 moles

Molarity of NaCl=moles of NaCl/Volume of solution=7.7x10^-3/0.1=0.077 M

Osmolarity of NaCl=2*0.077=154 Osm

Osmolarity of 0.9% NaCl

moles of NaCl=0.9/58.44=0.0154 moles

Molarity of NaCl=0.0154/0.1=0.154 M

Osmolarity of NaCl=2*0.154=0.308 Osm

Osmolarity of 5% urea:

moles of urea=5/60.06=0.083 moles

Molarity of urea=0.083/0.1=0.83 M

Osmolarity of urea=2*0.83=1.66 Osm

Osmolarity of 2% urea:

moles of urea=2/60.06=0.033 moles

Molarity of urea=0.033/0.1=0.33 M

Osmolarity of urea=2*0.33=0.66 Osm

Osmolarity of 5% glucose:

moles of glucose=5/180=0.0278 moles

Molarity of glucose=0.0278/0.1=0.278 M

Osmolarity of glucose=2*0.278=0.556 Osm

i would go with b

cytokinesis occurs after telophase so that eliminates d

and the entire reason for cytokinesis is for cellular reproduction

the match duh

explanation: