Question

$\huge\mathbb\underline{Question-}$

$<b>Calculate the weight of Na2CO3 required for the exact titration of 100 ml, decimolar HCl solution</b>$

☺☺☺☺☺☺​

Answer 1

$\huge\bold\red{ANSWER:-}$

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.

_________________________

Plz mark as brainlist.....

Answer 2

$</p><p></p><p><style></p><p>img{</p><p>width:353px;</p><p></p><p>}</p><p></p><p></style></p><p></p><p><img src="https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcSlbl2rJ9pGAsnSk95bAYSAWaVPIsQ-iIGyJA&usqp=CAU"</p><p></p><p>$