Hphys Lab Final Review

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Osmolarity Problems. 1. Nonionizing substances: One osm of nonionizing substance, such as glucose, 1 mmole of the substance. One mmole of glucose (C6H12O6) is 180mg. If 180mg. If 180mg of glucose is placed in a beaker and distilled water added to make one liter, the osmolarity of the solution is 1mosm/L A) How many mOsm will 1 gram of glucose yield? Show calculations. (1gC6H12O6/1)*(1000mg/1g)(1mOsm/180mgC6H12O6)=5.55mosm 2. Zing Substances. For ionizing substances, such as NaCl , 1mosm is 1mmole times the number of ions formed when each molecule dissociates. One mmole of NaCl is 58 mg, but when it dissociates, it yields 1 mmole of Na+ (23mg) and 1mmole of Cl ( 35mg). Therefore , 58 mg of NaCl is 2 mOsm of NaCl is put into a beaker and distilled water added to make 1 liter, the osmolarity is 2mOsm/l. A) How many mosm solute will 1 gram of NaCl yield? Show your calculations. (1gNaCl/1)*(1000mg/1gNaCl)(2/58mg)=34.5mOsm. 3. Mixed Solutions: If 1 mmole of glucose (180mg=1mOsm) and 1 mmole of NaCl (58mg=2mmOsm) are put into a beaker and distilled water added to make 1 liter, the osmolarity is 3 mOSm/L. OSMOSIS CALCULATIONS: 1. Calculate the number of grams of NaCl needed to prepare 100 mL of a 280mOsm.L Solution. (280mosm/l) *(1L/1000mL)*(100mm/1)*(29mg/1mosm)*(1g/1000mg)=0.812g 2. Calculate the number of grams of glucose needed to prepare 100mL of a 280 msm/L glucose solution. (280mOsm/L)*(1L/1000mL)(100/1)*(120mg/1mosm)*(1g/1000mg)=5.04g Based on your calculation a 0.812% NaCl solution (w/w) and a 5.04% glucose (w/w) solution are isotonic to normal human cells. In medical professions clinicians consider a 0.9% NaCl and a 5% glucose solution is isotonic to normal human

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