Question

Estimate the vapor pressure of seawater at 20°C given that the vapor pressure of pure water is 2.338 kPa at that temperature and the solute is largely Na⁺ and Cl^- ions, each present at about 0.50 mol/dm³.

Answer 1

Solution:

Estimate the vapor pressure of seawater

NaCl→Na+(aq)+Cl−(aq)

We have:

nNa+=0.5mol
nCl−=0.5mol
nNaCl=5mol

Formula PJ=XJP

Or Ppurewater=XJ(water)×PSeawater⇒PSeawater=XJ(water)Ppurewater

Since Ppurewater=2.338kPa

• Find XJ(H2O)

Formula XJ=nJn=nH2Ontotal

• Find nH2O

Formula nH2O=mH2OMH2O,MH2O=18g/mol

But mH2O=mseawater−mNaCl since mseawater=1000g

mNaCl=nNaCl×MNaCl since nNaCl=0.5mol,MNaCl=58.5g/mol

So, mNaCl=0.5mol×58.5g/mol

Therefore, mH2O=1000g−29.25g=970.75g

Then \(n_{H_2 O} = \frac{970.75 \, \text{g}}{18 \, \text{g/mol}} = 53.93 \, \text{mol}

• Find ntotal

Since \(n_{total} = n_{H_2 O} + n_{Na^+} + n_{Cl^-} = (53.93 + 0.5 + 0.5) \, \text{mol} = 54.95 \, \text{mol}

Thus, XJ(H2O)=53.93mol54.95mol=0.9817

We get \(P_{Seawater} = \frac{2.338 \, \text{kPa}}{0.9817} = 2.3815 \, \text{kPa}

Therefore, the vapor pressure of seawater is \(P_{Seawater} = 2.3815 \, \text{kPa}

Therefore, the vapor pressure of seawater is \(P_{Seawater} = 2.3815 \, \text{kPa}