17.1 g of Al2(SO4)3 (molar mass 342 g mol^-1) is completely precipitated as BaSO4 by BaCl2 solution (treat BaCl2 as having n=2). Select the correct statement(s).

The n-factor of Al2(SO4)3 in this reaction is 6

Its equivalent mass is 57 g mol^-1

The n-factor of Al2(SO4)3 in this reaction is 6

Its equivalent mass is 57 g mol^-1

600 mL of 0.50 M BaCl2 is required

JEE Main levelOne or more correct optionsDifficulty 2 of 3eq012

$17.1 g$ of $Al X_{2} (SO X_{4}) X_{3}$ (molar mass $342 g mo l^{- 1}$ ) is completely precipitated as $BaSO X_{4}$ by $BaCl X_{2}$ solution (treat $BaCl X_{2}$ as having $n = 2$ ). Select the correct statement(s).

Three

SO X_{4} X^{2 -}

ions each carry

2

charges, so

n = 6

and equivalent mass

= 342/6 = 57

. Equivalents

= 17.1/57 = 0.30

. A

0.50 M

BaCl X_{2}

solution is

1.0 N

, so volume

= 0.30/1.0 = 0.30 L = 300 mL

, hence option 4 (

600

mL) is wrong.