What are the molality and mole fraction of solute in a 35.9 percent by mass aqueous solution of formic acid (hcooh)?

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1- The molarity of HCOOH = 9.515 M.

2- The mole fraction of HCOOH = 0.18.

Explanation:

1- The molarity of HCOOH:

M = (10pd)/molar mass.

p is the percent by mass of HCOOH = 35.9 %.

d is the specific gravity of HCOOH = 1.22 g/cm³.

Molar mass of HCOOH = 46.03 g/mol.

∴ M = (10pd)/molar mass = (10)(35.9 %)(1.22 gcm³) / (46.03 g/mol) = 9.515 M.

2- The mole fraction of HCOOH:

The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water).

no. of moles of HCOOH = mass / molar mass = (35.9 g)/(46.03 g/mol) = 0.78 mol.

no. of moles of water = mass / molar mass = (64.1 g)/(18.0 g/mol) = 3.56 mol.

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.