1. Calculate the total osmolarity of a 250 mL solution consisting of 0.02 M CaCl

Question

1. Calculate the total osmolarity of a 250 mL solution consisting of 0.02 M CaCl2, 0.1 M NaCl and 0.05M dextrose. what percentage (%W/V) of NaCl must a solution contain to have an osmolarity of 200 mOsmol/L? What is the osmolarity of a solution containing 0.25% NaCl? 2, 3. Extra Strength Alka-Seltzer Effervescent Tablets contain 1,985 mg of sodium bicarbonate (NaHCO3, MW 84) per tablet. Calculate the osmolarity of a solution made by dissolving 2 tablets in 8 fluidounces of water (assume final volume of solution is 8 fluidounces). 4. Calculate the molality of a 750 ml solution containing 120g of glucose (MW 180). The density of the final solution 1.015. 5. What is the molarity of a 10% solution of calcium gluconate (Ci2H2Ca014, MW - 430)? What is the equivalent weight of calcium gluconate?

Explanation / Answer

1. total osmolarity = 0.02*3+0.1*2+0.05 = 0.31 M

    1 CaCL2 = 3 ions(Ca^2+,2CL-)

    1 NaCl =2 ions(Na+,Cl-)

    1 dextrose = 1 molecules

2. osmolarity = 2*molarity of naCl

molarity of naCl = 200*10^-3/2 = 0.1 M

molarity = (w/v%)*10/Mwt

0.1 = x*10/58.5

x = w/v% = 0.585%

3. 1 NaHCO3 = 2 ions

mass of NaHCO3 dissolved = 1985*2*10^-3 = 3.97 g

volume of solution = 8 fluidounces = 236.6 ml

osmolarity(C) = 2*molarity

                = 2*(3.97/84)*(100/236.6)

                = 0.04 M

4. molality(m) = (w/mwt)*(1000/Wt of solvent in g)

mwt = molarmass of glucose = 180 g/mol

Wt = mass of solvent = 750*1.015 - 120 = 641.25 g

w = mass of solute = 120 g

molality = (120/180)*(1000/641.25)

   = 1.04 m

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