The normal boiling point of methanol is 64.7 ∘ c and the molar enthalpy of vaporization is 71.8kj/mol . the value of δs when 1.35 mol of ch 3 oh(l) vaporizes at 64.7 ∘ c is j/k .

286 J/K

Explanation:

The molar Gibbs free energy for the vaporization (ΔGvap) is:

ΔGvap = ΔHvap - T.ΔSvap

where,

ΔHvap: molar enthalpy of vaporization

T: absolute temperature

ΔSvap: molar entropy of the vaporization

When T = Tb = 64.7 °C = 337.9 K, the reaction is at equilibrium and ΔGvap = 0.

ΔHvap - Tb . ΔSvap = 0

ΔSvap = ΔHvap/Tb = (71.8 × 10³ J/K.mol)/ 337.9 K = 212 J/K.mol

When 1.35 mol of methanol vaporizes, the change in the entropy is:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).