Question
Asked in: JEE Main-2019
A mixture of 100 m mol of 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 in resulting solution , respectively , are : (Molar mass of , are 74, 143 and 136 g , respectively ; )
A.
B.
C.
D.
Na2SO4 + Ca(OH)2 → CaSO4 + 2 NaOH
mmol of Na2SO4 = 2*1000/143
=13.98 m Mol
mmol of CaSO4 formed = 13.98 m Mol
Mass of CaSO4 formed = 13.98 × 10^-3 × 136 = 1.90 g
mmol of NaOH = 28 mmol
Ca(OH)2 → Ca2+ + 2 OH-
[OH–] = 28 mmol / 100ml=0.028 mol/0.1 L =0.28mol L-1
Study 40% syllabus and score up to 100% marks in JEE