The required amount of crystalline oxalic aci | vedclass

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Question

(B) The formula for normality is given by: $N = \frac{W 	imes 1000}{E 	imes V(mL)}$Given:Normality $(N)$ = $N/10 = 0.1 \, N$Equivalent weight $(E)$

Vedclass · Accepted Answer

$1.575$

Vedclass · Answer

(B) The formula for normality is given by: $N = \frac{W 	imes 1000}{E 	imes V(mL)}$Given:Normality $(N)$ = $N/10 = 0.1 \, N$Equivalent weight $(E)$ = $63$Volume $(V)$ = $250 \, mL$Substituting the values into the formula:$0.1 = \frac{W 	imes 1000}{63 	imes 250}$$0.1 = \frac{W 	imes 4}{63}$$W = \frac{0.1 	imes 63}{4} = \frac{6.3}{4} = 1.575 \, g$Therefore,the required amount is $1.575 \, g$.

The required amount of crystalline oxalic acid (equivalent weight = $63$) to prepare $250 \, mL$ of $N/10$ oxalic acid solution is: (in $, g$)

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