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Solve! - A mixture of 100 m mol of and 2 g of sodium sulphate was dissolved in water and the volume was made - Solutions - JEE Main

A mixture of 100 m mol of Ca(OH)_{2} and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL .The mass of calcium sulphate formed and the concentration of OH^{-} in resulting solution , respectively , are  : (Molar mass of  Ca(OH)_{2}  ,Na_{2}SO_{4} \: and \: \: CaSO_{4} are 74, 143 and 136 g mol^{-1} , respectively ; K_{sp} \: \: of\: \: Ca(OH)_{2}\: \: is\: \: 5.5\times 10^{-6})

  • Option 1)

    1.9\: g,\: \: 0.28\: mol\: L^{-1}

  • Option 2)

    13.6g,\: \: 0.28\: molL^{-1}

  • Option 3)

    1.9g,\: \: 0.14molL^{-1}

  • Option 4)

    13.6g,\: \: 0.14molL^{-1}

Answers (1)
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Mole Fraction -

\dpi{100} Mole\: Fraction= \frac{Moles\: o\! f\: solute}{Moles\: o\! f solute+Moles\: o\! f solvent}

-

As we have learned in mole concept .

The reaction is -

Ca(OH)_{2} + Na_{2}SO_{4} \rightarrow CaSO_{4} + 2 NaOH

 

 

 

Vapour Pressure -

It is defined as pressure exerted by vapours on liquid surface at equilibrium and condensation.

-

 

  

 

Raoult's Law -

The partial pressure of any volatile constituents of a solution at a given temperature is equal to the product of vapour pressure of pure constituent and its mole fraction in the solution.

P_{A}=P_{A}^{0}x_{A}

P_{B}=P_{B}^{0}x_{B}

 

 

- wherein

P_{A}^{0}  and P_{B}^{0} are vapour pressures of pure liquids.

P_{A}  and P_{B} are vapour pressures of A and B respectively.

 

 

 

Mathematical Expression -

\Delta T_{b}= K_{b}\: m

Unis of K_{b}=\frac{K-K_{g}}{mole}

\Delta T_{b}= Elevation\: in \: boiling\: point
 

- wherein

K_{b}= Boiling \: point \: elevation \: constant

m= molality

 

 


Option 1)

1.9\: g,\: \: 0.28\: mol\: L^{-1}

Option 2)

13.6g,\: \: 0.28\: molL^{-1}

Option 3)

1.9g,\: \: 0.14molL^{-1}

Option 4)

13.6g,\: \: 0.14molL^{-1}

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