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  • Question 1
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    Directions For Questions

    The vapour pressure of two pure liquids $$A$$ and $$B$$ which form an ideal solution are $$500$$ and $$800$$ torr respectively at $$300K$$. A liquid solution of $$A$$ and $$B$$ for which the mole fraction of $$A$$ is $$0.60$$ is contained in a cylinder closed by a piston on which the pressure can be varied. The solution is slowly vaporized at $$300K$$ by decreasing the applied pressure.

    ...view full instructions

    The composition of vapour when first bubble formed is:
    Solution

  • Question 2
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    A liquid A and B form an ideal solution. If vapour pressure of pure A and B are $$500N{ m }^{ -2 }$$ and $$200N{ m }^{ -2 }$$ respectively, the vapour pressure of a solution of A and B containing 0.2 mole fraction of A would be:
    Solution
    Vapour pressure of a solution containing A & B liquid is
    $$P_T=P_A+P_B$$
    $$P_A= X_A-P^o_A$$               $$X_A$$= mole fraction of A; $$P^o_A$$= Pure vapour pressure of A.
    $$\Rightarrow P_A= 0.2 \times 500 Nm^{-2}$$ & $$P_B= (1-0.2)\times 200 Nm^{-2}$$
    Thus $$P_T= 0.2 \times 500 Nm^{-2} + 0.8 \times 200 Nm^{-2}$$
    $$\Rightarrow P_T= 260 Nm^{-2}$$
    Thus total vapour pressure of solution is $$260 Nm^{-2}$$ .
  • Question 3
    1 / -0
    A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of $$290\ mm$$ at $$300\ K$$. The vapour pressure of propyl alcohol is $$200\ mm$$. If the mole fraction of ethyl alcohol is $$0.6$$, its vapour pressure (in $$mm$$) at the same temperature will be:
    Solution
    By Roult's law, 
    $${ P }_{ T }={ P }_{ A }^{ \circ  }{ x }_{ A }+{ P }_{ B }^{ \circ  }{ x }_{ B }$$

    $$\therefore \quad 290mm={ P }_{ A }^{ \circ  }\times 0.6+200\times \left( 1-0.6 \right) $$

                            $$={ P }_{ A }^{ \circ  }\times 0.6+80$$

            $$\therefore \quad { P }_{ A }^{ \circ  }=350\  mm$$

    Hence, the correct option is $$D$$
  • Question 4
    1 / -0
    A $$3.00\ L$$ oxygen gas is collected over water at $${27}^{o}C$$ when the barometric pressure is $$787\ Torr$$. If vapour pressure of water is $$27\ Torr$$ at $${27}^{o}C$$, moles of $$O_{2}$$ gas will be?
    Solution
    Given Volume (V)$$=3.00$$ L
    Temperature(T)$$=27^{0}C=300 K$$
    Barometric pressure$$=787$$ torr
    Vapour pressure of water$$=27$$ torr
    total pressure$$=787-27=760\quad{torr}=1\quad{atm}$$
    Applying in ideal gas equation
    PV=nRT
    $$n=\dfrac{PV}{RT}$$
    $$n=\dfrac{{1}\times{3}}{{0.0821}\times{300}}=0.122$$
  • Question 5
    1 / -0
    The vapour pressure of pure water is 760 mm at $$25^{ 0 }C$$.The vapour pressure of solution containing 1(m) solution of glucose will be:
    Solution
    Vapor pressure  
    $$\dfrac{P^{o}- P}{P^{o}}= \dfrac{W_{2}}{M_{2}} \times \dfrac{M_{1}}{W_{1}}$$

    $$P^{o}= 760\ mm$$

    $$M_{1} = 18$$

    Molality $$(m)= 1m= \dfrac{mole\ of\ solute}{(weight)\ kg\ of\ solvent }= \dfrac{W_{2}/M_{2}}{W_{i} m \ gm}$$

    $$\dfrac{760-P}{760} = 1 \times 10^{-3} \times 18$$

    $$P= 746.5\ mm$$

    Hence, the correct option is $$B$$
  • Question 6
    1 / -0
    Which of the following is least soluble?
    Solution
    Here the option (B)$$Sr{ F }_{ 2 }$$ is correct, Which is least soluble.
    Hydrogen energy decreases down the group, so solubility with the water decreases down the group.
    Solubility in the order $$Be>Mg>Ca>Sr>Ba$$
  • Question 7
    1 / -0
    Which of the following condition is not followed for an ideal solution?
    Solution
    An ideal solution satisfies the following conditions.
    1)There will be no change in volume on mixing two components.
    $$\triangle V_{mix}=0$$
    2)There will be no charge in enthalpy on mixing two components.
    $$\triangle H_{mix}=0$$
    So, the $$\Delta S_{mixing}=0$$ is not followed by the ideal solution.
  • Question 8
    1 / -0
    Which of the following pair will form an ideal solution?
    Solution
    The substances that have similar structures and polarities form nearly ideal solution benzene and toulene will form an ideal solution.
    Acetone, chloroform, chlorobenzene chloroethane ,water $$HCl$$ will form non-ideal solutions due to the difference of polarites between the two components.
  • Question 9
    1 / -0
    $$3$$ moles of $$P$$ and $$2$$ moles of $$Q$$ are mixed, what will be their total vapour pressure in the solution if their partial vapour pressures are $$80$$ and $$60$$ respectively?
    Solution
    Mole fraction of $$P=\dfrac { 3 }{ 3+2 } =\dfrac { 3 }{ 5 } $$

    Mole fraction of $$Q=\dfrac { 2 }{ 3+2 } =\dfrac { 2 }{ 5 } $$

    Hence,

    Total vapour pressure =(Mole fraction of $$P$$ $$\times $$ Vapour pressure of $$P$$) + (mole fraction of $$Q$$ $$\times$$ Vapour pressure of $$Q$$)

    $$=\left( \dfrac { 3 }{ 5 } ×80 \right) +\left( \dfrac { 2 }{ 5 } ×60 \right) =48+24=72\ torr$$
  • Question 10
    1 / -0
    How many grams of sucrose must be added to $$320g$$ of water to lower the vapour pressure by $$1.5mm$$ $$Hg$$ at $${25}^{o}C$$?
    (Given: The vapour pressure of water at $${25}^{o}C$$ is $$23.8mm$$ $$Hg$$ and molar mass of sucrose is $$324.3g/mol$$)
    Solution
    $$w_{1}=$$ Wsucrose = ?
    $$M_{1}=$$ Molecular weight $$+32,4.3\ g/mol$$
    $$w_{2}=320\ g$$
    $$M_{2}=H_{2}O=18\ g/mol$$
    $$p-ps=1.5\ mm$$
    $$p=23.8\ mm$$
    $$\dfrac{p-ps}{p}=\dfrac{W_{1}}{M_{1}}\times \dfrac{M_{2}}{W_{2}}$$
    $$\dfrac{1.5}{23.8}=\dfrac{W_{1}}{324.3}\times \dfrac{18}{320}$$
    $$W_{1}=363.36\ g$$
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