Concentration and Tonicity Tube and ContentsMolality Visual Appearance of RBCs E

Question

Concentration and Tonicity Tube and ContentsMolality Visual Appearance of RBCs Estimated RBC Diameter (um) How many in field? 1.) 10g/100ml NaCI 2.) 3.5g/100ml NaCI 3.) 0. 85g/100ml NaCI 4.) 0.45g/100ml NaCI 5. 0.2g/100ml NaCi lco 2 l2 Molality (m) is the number of moles of solute per 1000 kg of solvent. Osmolality (Osm) is the ucosc upon dissolution, there is one mole of glucose particles in the 1000 kg of water. (Im glucose solution = 1 Osm solution) When making a 1 m NaCl solution, upon dissolution, there are two moles of solute particles. (NaCl Na+ + CT) (1m NaCl solution-20sm solution), what will a 1m CaClz equal in Osm? (@wCr-v1,'t5:1 dl/l 62017 mstww dwn

Explanation / Answer

1) 10g/100ml

Molality = number of moles in a solute/ 1000kg of solvent  

=no. of moles = 10g * 1moles of NaCl/58.5 g = 10*0.01709= 0.1709

=0.170/0.1= 1.7 molality

Osmolality= Molality *2 for Nacl as Nacl dissociates into 2 moles

1.7*2=3.4

Milliosmolality = 3.4*1000= 3400

2) 3.5g/100ml

Molality = 0.59

Milliosmolality= 0.59*2*1000= 1190

3) 0.85g/100ml

Molality = 0.145

MilliOsm= 290

4) 0.45g/100ml

No. of moles = 0.45*0.01709 = 0.00769

Molality = 0.00769/0.1= 0.0769

MilliOsm= 0.0769*2*1000=153.81

5) 0.2g/100ml

No. of moles = 0.2*0.01709 = 0.003418

Molality = 0.003418/0.1= 0.03418

MilliOsm= 0.03418*2*1000=68.36

Osmomolality = 0.145*2= 0.290

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