Question

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawater contains only NaCl, estimate its molal concentration.

Answer 1

Answer:

Molal concentration of NaCl in Seawater = 1.6 moles of NaCl/kg of pure water!!!

Explanation:

Let the mole fraction of pure water be X, that of NaCl = 1 - X

We obtain the mole fraction of solvent from Raoult's law relation

Pressure of solution, P⁰ₛₐₗₜ = Vapour pressure of solvent, P⁰ₕ₂₀ × mole fraction of solvent, X

23.09 = 23.76 × X

X = 0.972

Using a mole basis of 1 mole,

Number of moles of pure water = 0.972

number of moles of NaCl in the seawater sampler = 1 - 0.972 = 0.028 moles

Molal concentration = number of moles of solute/mass of solvent in kg

Mass of solvent = number of moles × molar mass of pure water = 0.972 × 18 = 17.496g = 0.0175 kg

Molal concentration = 0.028/0.0175 = 1.6 moles of NaCl/kg of pure water!!!

Hope this Helps!!!!

Answer 2

Answer:

0.808  M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$  ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$  ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$$= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity =  $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$