The specific heat of a certain substance is $0.86 \,J \,g^{-1} \,K^{-1}$. Assuming ideal solution behaviour,the energy required (in $J$) to heat $10 \,g$ of $1 \,molal$ of its aqueous solution from $300 \,K$ to $310 \,K$ is closest to $.... \,J$
[Given: Molar mass of the substance $= 58 \,g \,mol^{-1}$; specific heat of water $= 4.2 \,J \,g^{-1} \,K^{-1}$]
[Given: Molar mass of the substance $= 58 \,g \,mol^{-1}$; specific heat of water $= 4.2 \,J \,g^{-1} \,K^{-1}$]
- A$401.7$
- B$424.7$
- C$420.0$
- D$86.0$
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Similar Questions
The values of $K_b$ and $K_f$ for water are $0.52 \, K \, kg \, mol^{-1}$ and $1.86 \, K \, kg \, mol^{-1}$ respectively. If the solution boils at $0.78 \, K$ above the boiling point of water,then the freezing point of the solution will be ........ $K$.
Medium
View Solution$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution?
Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$
(Assume molarity to be equal to molality) $..... ^oC$
Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$
(Assume molarity to be equal to molality) $..... ^oC$
Medium
View SolutionAt $300 \ K$,an ideal solution is formed by mixing $460 \ g$ of toluene with $390 \ g$ of benzene. If the vapour pressure of pure toluene and pure benzene at $300 \ K$ are $32 \ mm$ and $40 \ mm$ respectively,the mole fraction of toluene in the vapour phase is:
An aqueous solution of a non-volatile solute boils at $100.17^{\circ} C$. The temperature at which this solution will freeze (in $^{\circ} C$) is
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$
$K_{b}(H_2 O) = 0.512^{\circ} C \ kg \ mol^{-1}$,
$K_{f}(H_2 O) = 1.86^{\circ} C \ kg \ mol^{-1}$
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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