Question

normality of 0.05 m na2 co3 solution is​

Answer 1

Molarity = NO. OF MOLES OF SOLUTE / VOLUME OF SOLUTION IN LITRE

HERE

MOLARITY=0.5M

VOLUME of solution =100ml

Molecular mass of Na2C03=23*2+12+16*3=106

Using formula and substituting value we have

0.5= x*1000 / 106*100

X=0.5*106*100/1000

X=5.3g

Or

if we convert me to Litre then

100ml=0.1litre

Mass = moles * molecular mass

=(0.1litre*0.5moles/litre)* 106grams

Mass=5.3grams.

Answer 2

it neautralizes 20 ml of 0.049 n of hcl. the gram equivalent of 0.04 n hcl is 1.46