Question

Calculate the molarity of a solution containing 6.3g oxalic acid dihydrate in 200gms of solution​.

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Answer 1

Answer:

normality is no. of gram equivalents per unit volume in litres, for oxalic acid, no of gram equivalents will be the given mass divided by equivalent mass.equivalent mass is molecular mass/nfactor.oxalic acid is h2c2o4 so it dissociates as H+ and C2O4^2- ions.hence net positive or negative charge is 2.hence n factor is 2. its molecular mass is its equivalent mass is 90/2 which is no of gram equivalents will be 1.26/45 which is 0.028.now normality is 0.028/250 . 250 ml is 1/4th of a litre hence 0.028/0.25 which is 0.028 x 4 which gives answer to be 0.112N.But sometimes oxalic acid is taken as h2c204 along with 2h20.hence then molecular msass is 126 g and equivalent mass is 126/2 which is gram equivalents become 1.26/63 which is 0.02 and normality eventually 0.02x4 i.e 0.08N

Answer 2

Answer:

0.25 is the molarity of a solution containing 6.3g oxalic acid dihydrate in 200gms of solution.

Explanation:

Formula of oxalic acid = H2C2O4.2H2O

Molar mass of oxalic acid = 126

NOW

MOLARITY =

$\frac{ \frac{mass \: of \: solute \: in \: g}{molar \: mass} \times 1000}{volume \: of \: solution \: in \: mililitre}$

$\frac{ \frac{6.3}{126} \times 1000 }{200}$

$\frac{ \frac{6300}{126} }{200}$

$\frac{50}{200}$

$0.25$