What is the molar solubility, molar concentration, of oxygen gas (O_2), in water

Question

What is the molar solubility, molar concentration, of oxygen gas (O_2), in water, if its partial pressure equals to 167 mm Hg at 25 degree C? (Henry's Law Constant for O_2 = 0.00131 mol/L middot atm) a) 6.37 times 10^3 mol/L b) 2.88 times 10^-4 mol/L c) 0.0259 mol/L d) 0.327 mol/L e) 4.72 times 10^-5 mol/L (Sec. 11.4) A solution is prepared by dissolving 7.75 grams of glucose (C_6H_12O_6), a non- electrolyte solute, in 235 grams of water at 25 degree C. Calculate the new vapor pressure of the solution. a) 16.4 mm Hg b) 28.8 mm Hg c) 0.259 mm Hg d) 23.7 mm Hg e) 26.1 mm Hg

Explanation / Answer

19.

C    = KH*P

P   = 167mmHg = 167/760   = 0.219atm

KH   = 0.00131mole/L-atm

C   = KH*P

      = 0.00131*0.219 = 2.88*10^-4mole/L >>>>>>answer B

20. no of moles of glucose n solute = 7.75/180 = 0.043 moles

      no of moles of H2O     n solvent    = 235/18    = 13 moles

mole fraction of solute( glucose)   =   n solvent/n solute + n solvent

                                                      = 13/0.043+13    = 13/13.043 = 0.996

    vapor pressure of solution = mole fractrion of solute * vapor pressure of solvent

                                                = 0.996*23.8   = 23.7mmHg >>>>>>D

                                           

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