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Equilibrium

Question
CBSEENCH11006619
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What is the minimum volume of water required to dissolve lg of calcium sulphate at 298 K? [For calcium sulphate, Ksp is 9·1 × 10–6).

Solution
CaSO subscript 4 left parenthesis straight s right parenthesis space space space space rightwards harpoon over leftwards harpoon space space space space Ca to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis Let space the space solubility space of space salt space in space water space be space straight S space mol space straight L to the power of negative 1 end exponent therefore space space space space open square brackets Ca to the power of 2 plus end exponent left parenthesis aq right parenthesis close square brackets space equals space straight S space mol space straight L to the power of negative 1 end exponent open square brackets SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis close square brackets space equals space straight S space mol space straight L to the power of negative 1 end exponent Then comma space straight K subscript sp space equals space open square brackets Ca to the power of 2 plus end exponent left parenthesis aq right parenthesis close square brackets space open square brackets SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis close square brackets 9.1 space cross times space 10 to the power of negative 6 end exponent space equals straight S space cross times space straight S or space space straight S space equals space left parenthesis 9.1 space cross times space 10 to the power of negative 6 end exponent right parenthesis to the power of 1 divided by 2 end exponent space equals space 3.02 space cross times 10 to the power of negative 3 end exponent space mol space straight L to the power of negative 1 end exponent
Molar space mass space of space CaSO subscript 4 space equals space left parenthesis 40 plus 32 plus 64 right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 126 straight g space mol to the power of negative 1 end exponent therefore space space Mass space of space CaSO subscript 4 space equals space left parenthesis 3.02 space cross times space 10 to the power of negative 3 end exponent space mol space straight L to the power of negative 1 end exponent right parenthesis space cross times space left parenthesis 126 straight g space mol space straight L to the power of negative 1 end exponent right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.411 space straight g space straight L to the power of negative 1 end exponent
Thus, volume of water required to dissolve 0.411 g of  CaSO4= 1L
therefore space space space Volume space of space water space required space to space dissolve space 1.0 space straight g space of space space CasO subscript 4 space space equals space fraction numerator left parenthesis 1 straight L right parenthesis space cross times space left parenthesis 1.0 space straight g right parenthesis over denominator left parenthesis 0.41 space 1 straight g right parenthesis end fraction equals space 2.43 space straight L
 

Some More Questions From Equilibrium Chapter

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

What is a backward reaction?

What type of equilibrium is attained in a reversibe chemical change?

What is meant by a system at dynamic equilibrium?

discuss the effect of a catalyst on equilibrium ? 

At equilibrium, the mass of each of the reactants and products remains constant. Does it mean that the reaction has stopped? Explain.

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