IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

ayshwarya

1) The number of moles of Ca(OH)2 to be dissolved in 500 mL of solution to get a solution of pH=10, assuming complete dissociation of Ca(OH)2

(ans 2.5 x 10^-5)

2) The compressibility factor "z" is greater than unity . Then at NTP molar volume Vm in ltrs. (ans Vm>22.4 )

3) At low pressure ranges, the vander waal's equation reduces to

(ans z = 1--a/RTV)

4) The heat liberated due to the combustion of acetelene is 1305 kJ and the heats of formation of H2O (I) and CO2(g) are -285 and -393 kJ respectively. Find the bond energy of (c=c) . Given BE (C-H) = 415kJ (H-H) = 436kJ and the heats of atomizatin Cg (s) is 715 kJ. (ans 700 )

5) If bond moments for CH3OCH3 are mh-c=0.40D, mc-o = 0.7 D and dipole moment of CH3OCH3 is 1.30 D , then bond angle C-O-C in CH3OCH3 ( in degree unit and in 10^-1 order ) is (ans 1075 )

chimanshu_007

@ low pressure , Vander waal's =n is given by PV = RT

the =n u wrote is V W =n @ moderate temp.

the fact for that is at moderate temp. , V decreases.....so a/V² increases . But V is still large in comparison with b dat b can be neglected

so the =n becomes

(P +a/V²).(V - 0) = RT

or PV + a/V = RT

PV/RT = 1 - (a/RTV)

or Z = 1 - (a/RTV)

dats ur answer no. 3

chimanshu_007

ans 2)Z = PV/nRT

V/n = V_m ,

Z > 1

PV_m/RT > 1

or V_m > 22.4

karthik2007

4) is easy. Use the fact that BE(reactants) - BE(products) = Enthalpy change.

Moreover, remember that:

Combustion is the reaction where adequate amount of oxygen reacts with one mole of the substance to give the required product. Try solving it using these hints. Else I'll solve it for you

adi_dev

ph=10
[H+]=10^(-10)M
[OH-]=10^(-4)M

It means in 1L sol. moles of [OH-]=10^(-4)
in .5L sol moles of [OH-]=5 X 10^(-5)
therefore,moles to be dissolved = 5 X 10^(-5)/2 = 2.5 X 10^(-5)

plz.. do rate me

ayshwarya

chimansu , how could volume decrease with decrease please explain

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