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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)

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

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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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