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13. What is the concentration of a 0.500 M NaCl solution in terms f a. mass perc

ID: 716878 • Letter: 1

Question

13. What is the concentration of a 0.500 M NaCl solution in terms f a. mass percent b. mole fraction (assume the density of the solution is 1.00 g/mL, show your work) 14. Which aqueous solution would have the highest boiling point? a. 0.025 m CaCl b. 0.030 m NaCl c. 0.100 m glucose (C&H1206) d. 0.030 m Fe(NO)s 15. Which aqueous solution would have the highest melting point? a. 0.025 m CaCl2 b. 0.030 m NaCl c. 0.100 m glucose (CsH1206) d. 0.030 m Fe(NO)s 16. The vapor pressure of pure ethanol at 25°C is 59mM Hg. What is the vapor pressure of a solution composed of 4.630 g of tannic acid (Cr&Hs20%, molar mass 1701.19 g/mol) is dissolved in 50.00 mL of ethanol (C2HsO)? Show your work. 17. The osmotic pressure of an aqueous solution containing an g of an unknown peptide is 0.6117 atm. To make the solution, 0.1054 g of the protein was dissolved in enough water to make 2.000 mL. What is the molar mass of this peptide? Show your work.

Explanation / Answer


13) concentration of NaCl = 0.5 M

   a) Molarity(M) = %bymass*d*10/Mwt

      %bymass = x

       d = density = 1.00 g/ml

       M.wt = molarmass of NaCl = 58.5 g/mol

        0.5 = x*1*10/58.5

     %bymass = x = 2.925%

   b) 0.5 mol NaCl present in 1 L solution.

    molefraction of NaCl = mol of NaCl/total moles

   total moles = moles of NaCl + moles of water

moles of water = (1000-0.5*58.5)/18 = 53.93 mol

  
   molefraction of NaCl = 0.5/(53.93+0.5)

                        = 0.0092

14) greater the no of ions, greater the boiling point of solution.

     answer: d.0.030 m Fe(NO3)3

15) greater the no of ions, LOWEST the freezing point of solution.                

    answer: b.0.03 m NaCl

16)

raoults law

P0-P/P0 = i*Xsolute

P0-P/P0 = n2 / n1+n2

i = vanthoffs factor of solute(tannic acid) = 1

P = vapor pressure of solution = ?

p0 = vapor pressure of pure ethanol at this temperature = 59 mmhg

n1 = no of mol of solvent = (50*0.789/46) = 0.858 mol

n2 = no of mol of solute particles(KCl) = (4.63/1701.19) = 0.00272 mol

(59-x)/59 = 1*(0.00272/(0.00272+0.858))

P = vapor pressure of solution = 58.8 mmhg

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