Calculate the osmolarity of a solution which contains 180 gm of glucose per dl, 117 gm of Nacl per dl and 56 gm of BUN per dl-

Ans. is 'd' i.e., 70 osmol/L * One mole of osmotically active particles is called one osmole.* In case of non-dissociated solutes, one gram molecular weight of any substance contain similar number of osmotically active molecules (osmoles), thus a molar solution of glucose contains 1 osmole.* On the other hand, in case of dissociated solutes, one gram molecular weight of any substance contain the number of osmotically active molecules (osmoles) equal to the number of dissociated molecules, e.g., a molar solution of NaCl contains 2 osmoles (1 mole of Na+ and 1 mole of Cl-).* The osmolar concentration of a solution in osmole/litre is called osmolarity. When expressed in osmole/Kg of solution is called osmolality.* Normal serum osmolality is about 290 mOsm/kg.* A mole (gram molecular weight) is molecular weight of a substance in grams, e.g., molecular weight of glucose is 180, therefore one mole of glucose has 180 gm; and molecular weight of NaCl is 58-5 gm, therefore one mole of NaCl contains 58-5 gm of NaCl.1 mole of glucose = 1 gram molecular weight of glucose = 180 gm of glucose = 1 osmole of glucose1 mole of NaCl = 1 gram molecular weight of NaCl = 58[?]5 gm of NaCl = 2 osmoles of NaClComing back to question* The given solution contains:-i) 180 gm glucse per dl = 1 osmole/dl =10 osmole/Lii) 117 gm of Nacl per dl = 4 osmole/dl = 40 osmole/Liii) 56 gm if BUN per dl = 2 osmole/dl = 20 osmole/L(28 gm of BUN has one osmole)* Thus osmolarity of given solution is 70 osmole/L