The mole fraction of KCl in the solution is 0.1051
calculation
mole fraction of KCl in solution = moles of KCl / total number of moles(moles of KCl +moles of H2O)
moles=mass/molar mass
mass of KCl=32.7g
molar mass of KCl= 39 +35.5
moles of KCl is therefore= 32.7g/74.5 g/mol=0.439 moles
find the moles of H2O= mass of H2O/molar mass
mass of H2O=100-32.7=67.3g
molar mass of H2O=( 1 x2) +16=18 g/mol
moles = 67.3/18 =3.739 moles
total moles=3.739+0.439=4.178 moles
mole fraction is therefore=0.439/4.178=0.1051
Answer:
The MO method for N2+ gives the bond order equal to 2.5. But first, we look at the diagram of molecular orbitals for N2 (the bond order for the nitrogen molecule is 3). the N2+ molecule). That is, the bond order for N2+ is 2.5.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
M(C2H2O)= 12.0*2 +1.0*2 +16.0 = 42 g/mol is a molar mass for empirical formula.
120.6g/mol/42g/mol ≈ 3
So, empirical formula should be increased 3 times,
and molecular formula is C6H6O3.
Answer is D.
