A solution of sulphuric acid having 0.49 gram of it dissolved in 250 cm3 of solution will have its normality -(Equivalent mass of sulphuric acid = 49)
Question
A solution of sulphuric acid having 0.49 gram of it dissolved in 250 cm3 of solution will have its normality -(Equivalent mass of sulphuric acid = 49)
Solution
To find the normality of the solution, we need to use the formula:
Normality = Number of gram equivalents / Volume of solution in liters
First, we need to find the number of gram equivalents. Given that the equivalent mass of sulphuric acid is 49 and we have 0.49 grams of it, the number of gram equivalents is:
0.49 g / 49 g/equivalent = 0.01 equivalents
Next, we need to convert the volume of the solution from cm3 to liters. Since 1 cm3 is equal to 0.001 liters, 250 cm3 is equal to:
250 cm3 * 0.001 L/cm3 = 0.25 L
Now we can find the normality:
Normality = 0.01 equivalents / 0.25 L = 0.04 N
So, the normality of the solution is 0.04 N.
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