ِAnswer:
1- The molarity of HCOOH = 9.515 M.
2- The mole fraction of HCOOH = 0.18.
Explanation:
<em>1- The molarity of HCOOH:</em>
- We can calculate the molarity of HCOOH using the relation:
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.
<em>2- The mole fraction of HCOOH:</em>
- We can suppose that we have a 100 g solution, that contains 35.9 g of HCOOH and 64.1 g of water.
<em>The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water).</em>
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.
- The mole fraction of HCOOH = (no. of moles of HCOOH) / (no. of moles of HCOOH + no, of moles of water) = (0.78 mol) / (0.78 mol + 3.56 mol) = 0.18.
the 2nd one is correct
The option that would most likely increase the rate for a more contentrated potassium hydrogene
:) hope this helps
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