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(3) How much gram of sodium carbonate \( \left( Na _{2} CO _{3}\right) \) dissolved in \( 200 ml \) to get a noumality of \( 0.5 N \). |
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Answer» Since normality is defined as Normality= number of gram equivalent / Volume (l) Hence 0.5 = Ñ/0.2 where Ñ is number of gram equivalent solving we get Ñ= 0.1 Also number of gram equivalent = Mass/Eq mass and Eq mass = molecular mass / valence For given compound molecular mass is 106 and valence is 2 So Eq mass is 53 substituting know data we get mass as 0.1× 53 gm= 5.3 gm |
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