Question

1. A solution that contains 90 moles of water and 1 mole of MgCl2 will have what partial pressure? The vapor pressure of pure water is 17.5 mmHg.

At standard temperature and pressure, water has a vapor pressure of 17.5 mmHg, and ethanol has a vapor pressure of 43.7 mmHg. What is a possible vapor pressure of an ethanol/water mixture.

Nitrogen gas has a Henry's Law constant of 0.00061 mole/L•atm. At a concentration of 0.1 mole/L, what is the partial pressure of the nitrogen above water, assuming it is an sealed container?

How much heat is required to heat 100g of water vapor from 101-200ºC. The specific heat of water vapor is 1.996 J/g•ºC.

How much heat is released by 10,000g of 0º water as it freezes? The enthalpy of fusion for H2O is 333.55 J/g.

Osmotic pressure = iMRT. What is the osmotic pressure of a 0.8 M solution of NaCl? Assume a van t'Hoff constant of 1.9

For a certain reaction, the change in enthalpy is -180 kJ/mole. Calculate the change in entropy of the surroundings at 300K. ΔS = ΔH(system)/T

Answer 1

1) No . Of moles of MgCl2 = 1 mol

No of moles of water = 90 mol

Vapour pressure of water p_o = 17.5 mm of Hg

From Henry's law

partial pressure = p^o * X( mole fraction)

Mole fraction X = no of moles of water/ total number of moles

X = 90/ (90 + 1)= 90/91

Partial pressure = 17.5 * 90/91

17.307 mm of Hg

Vapour pressure of solution ps= ?

From Raoult's law for dilute solutions po-p_s /p_o =

no of moles of MgCl₂/ no of moles of water

Vapour pressure of solution p_s = p_o - ( No of moles of MgCl2/ no of moles of water * p_o)

_{= 17.5 - ( 1/90 * 17.5)}

_{ps = 17.305 mm of Hg}

2) Ethanol and water mixture shows positive deviation from Raoults law, hence P_total > P_A + P_B

So that have minimum boiling azeotropic

Hence mixture have more vapour pressure

3)

Molarity is 0.1 means 0.1 moles of solute present in 1000 ml of water

no of moles of solute is 0.1 mol

No of moles of solvent water = 1000/18 =55.5

Henry's constant K_H = 0.00061 mol / lit atm

From Henry's law P = K_H * X

P = 0.00061 * 0.1/(0.1+ 55.5)

=1.097 * 10^-6 atm

4) mass of water = 100 gr

Specific heat of water= 1.996 j/gr ℃

Difference in Temperature = 200 -101= 99℃

Amount of heat q = m s ∆T

= 100 * 1.996 * 99

= 19760.4 J or 19.7604 KJ

5) vant hoff factor i = 1.9

Temperature 25℃ or 298k

Universal gas constant R = 0.0821 lit atm /mol-k

Osmotic pressure = i MRT

= 1.9 * 0.8* 0.0821 * 298

37.188 atm

6) ∆S_surrounding = ∆H/T or -∆H_reaction/ T

= 180 KJ/ 300

180* 1000/ 300J

600 J/mol or 0.6 KJ/mol

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