Please sir solve this...
6.3 gm of oxalic acid (eq. Wt.=63) is dissolved in 500 ml solution what will be the normality of the solution?
oxalic is monoprotic, so
N=moles/volume=(6.3/63)/.5
I don't believe H2C2O4 has an equivalent weight of 63. The molar mass is 90 and it is diprotic; therefore, the equivalent weight is 45.
Sure! To find the normality of a solution, we need to know the equivalent weight (eq. Wt.) and the number of equivalents present in the solution.
In this case, we are given that the equivalent weight of oxalic acid is 63.
Now, we need to find the number of equivalents (eq) present in the solution using the formula:
Number of equivalents (eq) = mass of substance (in grams) / equivalent weight
Given that the mass of oxalic acid is 6.3 gm, and the equivalent weight is 63, we can substitute these values into the formula:
Number of equivalents (eq) = 6.3 gm / 63
Number of equivalents (eq) = 0.1
Since we know that:
Normality (N) = Number of equivalents (eq) / Volume of solution (in liters)
Given that the volume of the solution is 500 ml (or 0.5 liters), we can substitute these values into the formula:
Normality (N) = 0.1 eq / 0.5 L
Normality (N) = 0.2 N
Therefore, the normality of the solution is 0.2 N.