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Why is L.P.G used as domestic fuel, mixe...

Why is L.P.G used as domestic fuel, mixed with a strong smelling volatile liquid like mercaptan?

A

For better combustion

B

For good smell

C

For detection of gas leakage

D

For high calorific value

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Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B) where P_(A)^(@) and P_(B)^(@) are vapour pressures of pure components A and B and X_(A), X_(B) are their mole fractions in mixture. For solute-solvent system, the relatio becomes P_(M) = P_(A)^(@).X_(A) where B is non-volatile solute. A mixture of two volatile liquids A and B 1 and 3 moels respectively has a V.P of 300 mm at 27^(@)C . IF one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27^(@)C . The vapour pressure of A is:

Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B) where P_(A)^(@) and P_(B)^(@) are vapour pressures of pure components A and B and X_(A), X_(B) are their mole fractions in mixture. For solute-solvent system, the relatio becomes P_(M) = P_(A)^(@).X_(A) where B is non-volatile solute. If M is mol.wt. of solvent, K_(b) is molal elevation constant and t_(b) is its boiling point, P^(@) is its vapour pressure ta temperature T and P_(S) is vapour pressure of non-volatile solute in it at T K , then:

Vapour pressure of a solvent is the pressure exterted by vapour when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent of nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two valatile liquids is given by P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B) where P_(A)^(@) and P_(B)^(@) are vapour pressures of pure components A and B and X_(A), X_(B) are their mole fractions in mixture. For solute-solvent system, the relatio becomes P_(M) = P_(A)^(@).X_(A) where B is non-volatile solute. The amount of solute ("mol. wt. 60") required to dissolve in 180 g water to reduce the vapour pressure to 4//5 of the pure water: