NEET Chemistry Solutions MCQs

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    NEET Questions / Chemistry / Solutions

    5.

    All form ideal solution except

    A

    C6H6  and  C6H5NH2{C_6}{H_6}\,\,and\,\,{C_6}{H_5}N{H_2}

    B

    C6H6  and  C6H5I{C_6}{H_6}\,\,and\,\,{C_6}{H_5}I

    C

    C6H5Cl  and  C2H5Br{C_6}{H_5}Cl\,\,and\,\,{C_2}{H_5}Br

    D

    C6H5I  and  C6H5OH{C_6}{H_5}I\,\,and\,\,{C_6}{H_5}OH

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    6.

    Which of the following solutions has the highest normality?

    A

    6β€…β€Šgβ€…β€Šofβ€…β€Šβ€…β€ŠNaOH/1006{\rm{ }}\;g\;{\rm{ }}of\;\;{\rm{NaOH}}/100 mL{\rm{mL}}

    B

    0.5β€…β€ŠMβ€…β€ŠH2SO40.5\;{\rm{M}}\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}

    C

    N phosphoric acid

    D

    8β€…β€Šgβ€…β€Šofβ€…β€ŠKOH/L8{\rm{ }}\;g\;{\rm{ }}of\;{\rm{ }}KOH/L

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    7.

    A solute when distributed between two immiscible phases remains associated in phase II and dissociated in phase I. If Ξ±is the degree of dissociation and n is the number of molecules associated then :

    A

    K=cIcIIK = \frac{{{c_{\rm{I}}}}}{{{c_{{\rm{II}}}}}}

    B

    K=cIcII(1βˆ’Ξ±)nK = \frac{{{c_{\rm{I}}}}}{{\sqrt[n]{{{c_{{\rm{II}}}}\left( {1 - {\rm{\alpha }}} \right)}}}}

    C

    K=cIcII(1βˆ’Ξ±)K = \frac{{{c_{\rm{I}}}}}{{{c_{{\rm{II}}}}\left( {1 - {\rm{\alpha }}} \right)}}

    D

    K=cI(1βˆ’Ξ±)cIInK = \frac{{{c_{\rm{I}}}\left( {1 - {\rm{\alpha }}} \right)}}{{\sqrt[n]{{{c_{{\rm{II}}}}}}}}

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    8.

    If P0β€…β€Šandβ€…β€ŠPs{{P_0}\;and\;{P_s}} are the vapour pressure of solvent and solution respectively and N1β€…β€Šand N2{{N_1}\;and\,{N_2}} are the mole of solute and solvent then :

    A

    (P0βˆ’Ps)/P0=N1/(N1+N2)\left( {{P_0} - {P_s}} \right)/{P_0} = {N_1}/\left( {{N_1} + {N_2}} \right)

    B

    (P0βˆ’Ps)/Ps=N1/N2\left( {{P_0} - {P_s}} \right)/{P_s} = {N_1}/{N_2}

    C

    Ps=P0.N2/(N1+N2){P_s} = {P_0}.{N_2}/\left( {{N_1} + {N_2}} \right)

    D

    All of the above

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    9.

    If P0β€…β€Šandβ€…β€ŠPs{{P_0}\;and\;{P_s}} are the vapour pressure of solvent and solution respectively and N1β€…β€Šand N2{{N_1}\;and\,{N_2}} are the mole of solute and solvent then :

    A

    (P0βˆ’Ps)/P0=N1/(N1+N2)\left( {{P_0} - {P_s}} \right)/{P_0} = {N_1}/\left( {{N_1} + {N_2}} \right)

    B

    (P0βˆ’Ps)/Ps=N1/N2\left( {{P_0} - {P_s}} \right)/{P_s} = {N_1}/{N_2}

    C

    Ps=P0.N2/(N1+N2){P_s} = {P_0}.{N_2}/\left( {{N_1} + {N_2}} \right)

    D

    All of the above

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