Benzene and naphthalene form an ideal solution at room temperature. For this process,the true statement$(s)$ is (are)
$(A)$ $\Delta G$ is positive
$(B)$ $\Delta S_{\text{system}}$ is positive
$(C)$ $\Delta S_{\text{surroundings}} = 0$
$(D)$ $\Delta H = 0$

  • A
    $(A, B, C)$
  • B
    $(A, B, D)$
  • C
    $(A, C, D)$
  • D
    $(B, C, D)$

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At $T(K)$,the vapour pressure of pure benzene (molar mass $= 78 \ g \ mol^{-1}$) is $0.85 \ bar$. When $2.0 \ g$ of a non-volatile,non-electrolyte solute is added to $39 \ g$ of benzene,the vapour pressure of the solution at $T(K)$ is $0.83 \ bar$. The elevation in boiling point (in $K$) of the same solution is: ($K_b$ of benzene is $2.6 \ K \ kg \ mol^{-1}$)

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When $1 \ g$ each of compounds $AB$ and $AB_2$ are dissolved in $15 \ g$ of water separately,they increase the boiling point of water by $2.7 \ K$ and $1.5 \ K$ respectively. The atomic mass of $A$ (in $amu$) is $........ \times 10^{-1}$ (Nearest integer). (Given: Molal boiling point elevation constant $K_b = 0.5 \ K \ kg \ mol^{-1}$)

For a dilute solution containing $2.5 \ g$ of a non-volatile non-electrolyte solute in $100 \ g$ of water,the elevation in boiling point at $1 \ atm$ pressure is $2^{\circ} C$. Assuming the concentration of solute is much lower than the concentration of solvent,the vapour pressure ($mm$ of $Hg$) of the solution is (take $K_{b}=0.76 \ K \ kg \ mol^{-1}$)

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