A mixture of 2g of sodium sulphate and 100 m mol of Ca(. O H .)2 was dissolved in water and the volume was made up to 100 mL. What are the mass of calcium sulphate formed and the concentration of OH- in resulting solution, respectively? Given:- (Molar mass of Ca(O H)2, Na2SO4 and CaSO4 are 74, 143 and 136g mol- 1, respectively; Ks p of Ca(O H)2is 5.5× (10)- 6 )

Question

A mixture of 2g of sodium sulphate and 100 m mol of Ca(. O H .)2 was dissolved in water and the volume was made up to 100 mL. What are the mass of calcium sulphate formed and the concentration of OH- in resulting solution, respectively? Given:- (Molar mass of Ca(O H)2, Na2SO4 and CaSO4 are 74, 143 and 136g mol- 1, respectively; Ks p of Ca(O H)2is 5.5× (10)- 6 ) (A) 1.9 g, 0.14 mol L- 1 (B) 13.6 g, 0.28 mol L- 1 (C) 1.9 g, 0.28 mol L- 1 (D) 13.6 g, 0.14 mol L- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-xndzikxmpszz9xyr.jpg" /> Mass of CaSO4=(14 × 136/1000)=1.9 g CaSO4=(14 × 136/1000)=1.9 g [O H-]=(28/100)=0.28 M

$1.9 g, 0.14 m o l L^{- 1}$

$13.6 g, 0.28 m o l L^{- 1}$

$1.9 g, 0.28 m o l L^{- 1}$

$13.6 g, 0.14 m o l L^{- 1}$

Mass of $C a S O_{4} = \frac{14 \times 136}{1000} = 1.9 g$
$C a S O_{4} = \frac{14 \times 136}{1000} = 1.9 g$
$[O H^{-}] = \frac{28}{100} = 0.28 M$

1.9g,0.14molL−1

13.6g,0.28molL−1

1.9g,0.28molL−1

13.6g,0.14molL−1

Mass of CaSO4​=100014×136​=1.9g CaSO4​=100014×136​=1.9g [OH−]=10028​=0.28M

$1.9 g, 0.14 m o l L^{- 1}$

$13.6 g, 0.28 m o l L^{- 1}$

$1.9 g, 0.28 m o l L^{- 1}$

$13.6 g, 0.14 m o l L^{- 1}$

Mass of $C a S O_{4} = \frac{14 \times 136}{1000} = 1.9 g$
$C a S O_{4} = \frac{14 \times 136}{1000} = 1.9 g$
$[O H^{-}] = \frac{28}{100} = 0.28 M$