The following data are for water: density of liquid 0.997 g cm^-3 standard boili

Question

The following data are for water: density of liquid 0.997 g cm^-3 standard boiling point 372.78 K. at 1 bar DeltavapHdegree = 40.016 kJ mol" Assume the density and enthalpies are constant over the temperature range of interest. Calculate the temperature at which the equilibrium vapor pressure of pure liquid water is 0.835 bar. How many grams of salt (NaCl) per liter of water would be required to raise the boiling point of water at 0.835 bar to 100.0degreeC? Assume that NaCl is completely dissociated, i.e. 2 moles of solute per mole of NaCl.

Explanation / Answer

a)

we know that

ln (P2/P1) = ( dH /R) ( 1/ T1 - 1/T2)

given

P1 = 1

T1 = 372.78

P2 = 0.835

dH = 40.016 x 1000

R = 8.314

so

ln ( 0.835 / 1 ) = ( 40.016 x 1000 / 8.314 ) ( 1/ 372.78 - 1/T2)

T2 = 367.645

so

the temperature is 367.645 K

b)

now

we know that

elevation in boiling point is given by

dTb = i x kb x m

now

dTb = 373 - 367.645

dTb = 5.3546

now

5.3546 = 2 x 0.512 x m

m = 5.23

now

molality = moles of NaCl / mass of water (kg)

we know that

1 L of water = 1 kg

so

5.23 = moles of NaCl / 1

moles of NaCl = 5.23

now

mass = moles x molar mass

so

mass of NaCl = 5.23 x 58.44

mass of NaCl = 305.6

so

305.6 grams of NaCl is required

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.