Atomic weights: Na 23.0, K 39.1, Cl 35.5, O 16.0, H 1.0, C12.0. 1. If a solution

Question

Atomic weights: Na 23.0, K 39.1, Cl 35.5, O 16.0, H 1.0, C12.0.
1. If a solution contains 200 mg NaCl/dL, what is its concentrationin g/L? What is its molarity?
2. Mix 2 liters of 3 M NaCl with 1 liter of 6 M glucose. What isthe total concentration of the mixed
solution? What is the concentration of glucose in the mixedsolution? What is the concentration of NaCl?
3. Calculate the osmolarity of 300 mL of a 0.9% NaCl solution. Thedissociation constant for NaCl is
1.8. (osmolarity = molarity × dissociation constant)
4. Using the dissociation constant, calculate the actual osmolarityof 5% dextrose in 0.45% saline.
5. The freezing point depression of an ideal solution is 1.86degrees C/osmole/kg water.
A plasma sample has a freezing point depression of 0.525 degrees C.What is the osmolality (mosmoles/kg water) of the plasma sample?

Explanation / Answer

Sorry, I can only do some of them... I've not dealt with osmolaritybefore, so I'm not sure how to do Question 4, and I'm not confidentof my answer for Question 5, either... = ( 1. 'dL' = decilitre = 10-1 L (according toWikipedia...)      200 mg NaCl / dL        = 200 mg NaCl / 0.1 L        = 0.2 g NaCl / 0.1 L    = 2 g / L 2. First calculate the moles of NaCl and glucose in thesolution: moles (n) = concentration (c) * volume (v) n(NaCl) = 3 M * 2 L = 6 mol    n(glucose) = 6 M * 1 L = 6 mol Then calculate the new concentrations:     v(total) = 2 L + 1 L = 3 L    c(NaCl) = n(glucose) = 6 mol / 3L = 2 mol / L = 2M The final concentration of each species is the same, since there isthe same number of moles of each, in the same volume. 3. 0.9% NaCl usually means you have 0.9% NaCl by mass, diluted insolvent (water). We have a 300 mL solution, so:     mass(NaCl) = 0.009 * 300 = 2.7 g NaCl Next calculate the moles of NaCl, then divide by volume (0.3 L) toget concentration (molarity), which can then be used to calculateosmolarity:     n(NaCl) = mass / molar mass                 = 2.7 g / 58.44 g.mol-1                 = 0.04620 mol    c(NaCl) = 0.04620 mol / 0.3 L                 = 0.1540 M osmolarity = 0.1540 * 1.8 = 0.2772 (not sure what units thisis meant to be in...) 4. ??? 5. Osmolality is osmoles of solute perkilogram of solvent (different fromosmolarity = osmoles of solute per litreof solution). We want do do a calculation that will give us unitsof osmoles / kg water. The only way I can see to do this is todivide the freezing point depression of the sample by the freezingpoint depression of the ideal solution:     0.525 oC / 1.86(oC/osmoles/kg) = 0.2823 osmoles / kg Side note:    oC / osmoles /kg = oC / (osmoles . kg-1) =oC.(osmoles.kg-1)-1 =oC.osmoles-1.kg oC / oC.kg.osmoles-1 =osmoles / kg

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