Question

( B a(O H)_{2} max =342 y )
( 171 g=1 ) mole ( therefore 3429 ) is ( x Rightarrow x=2 ) moles
( 100 % rightarrow 2 ) moles
( 20 % longrightarrow 20 frac{x 2}{100}=0.4 ) moles
Ba ( (text { on }) ), rotal moles ( =2 times 0.4=0.8 ) ( H N O_{3} ) Molaity ( =2 M ) Total moles is ( M=n times frac{1000}{v} ) ( n=2 times frac{1200}{1000}=2 cdot 4 ) moles
Total man moles aud ( a=2.4-0.8=1.6 ) Tocal volume of ( B a(O M)_{2} ) is ( d=frac{w}{v} )
( V=frac{342}{0.57}=600 mathrm{ml} )
HNO ( _{3} ) uolume is 1200 Total nolume ( =600+1200=1800 )
( M=n cos frac{1000}{v}=1.6 times frac{1000}{1800} )
( M=0.88 mathrm{M} )

# 342 g of 20% by mass of Ba(OH)2 solution (sp. gr. 0.57) is reacted with 1200 mL of 2 M HNO . If the final density of solution is same as pure water then molarity of the ion in resulting solution which decides the nature of the above solution is: (a) 0.25 (b) 0.5 M (C) 0.888 M. (d) None of these

Solution