JEE Mains 04-Sep-2020 Shift 2 Solved Paper

Molar mass of Na2CO3.xH2O
⇒23×2+12+48+18x
⇒46+12+48+18x
⇒(106+18x)
Eqwt =

M

2

=(53+9x)
As nfactor in dissolution will be determined from net cationic or anionic charge; which is 2 so
Eqwt =

M

2

=53+9x
Gmeq=

wt

Eqwt

=

1.43

53+9x

Normality =

Gmeq

Vlitre

Normality =0.1=

1.43

53+9x 0.1

As volume =100 ml
=0.1 Litre
So 10−2=

1.43

53+9 x

53+9 x=143
9x=90
x=10.00