Asolution is 40.0% by volume benzene (c6h6) in carbon tetrachloride at 20c. the vapor pressure of pure benzene at this temperature is 74.61 mmhg and its density is 0.87865 g/cm3; the vapor pressure of pure carbon tetrachloride is 91.32 mmhg and its density is 1.5940 g/cm3. if this solution is ideal, its total vapor pressure at 20c is

84.30 mm Hg

Explanation:

In 100 cm³ of solution we have: 40 cm³ C6H6 and 60 cm³ CCl4. Given the densities we can calculate their masses and number of moles, and since by Raoult´s law

Ptotal = XAPºA + XBPºB

where XA= mol fraction =na/(na +nb) and PºA vapor pressure pure of pure component A

m C6H6 = 40 cm³ x 0.87865 g/cm³ = 35.146 g

mol C6H6 = 35.146 g/ 78.11 g/mol = 0.45 mol

mass CCl4 = 60 cm³ x 1.5940 g/cm³ = 95.640 g

mol CCl4 = 95.640 g / 153.82 g/mol = 0.62 mol

mol tot = 1.07

XC6H6 = 0.45/ 1.07 = 0.42      XCCl4 = 0.62/1.07 =0.58

Ptot (mmHg) = 0.42 x 74.61 + .58 x 91.32 = 84.30 mmHg

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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