Question

How many grams of anhydrous Na2SO4 is required to prepare 200 ml of 0.2 M Na2SO4 solution?

Answer 1

ans. 0.568 may be

Explanation:

If 0.2⋅L of a 0.2mol⋅L−1 solution are required, this is a molar quantity of 0.2⋅L×0.2⋅mol⋅L−1=0.04mol.

Now, we simply multiply this molar quantity by a molar mass to get the mass:

0.04mol×142g⋅mol−1 = ??⋅g

Answer 2

Answer:

molarity = weight of Na2SO4 * 1000 / molecular weight of Na2SO4 * vol.(ml)

0.2 = weight of Na2SO4 * 1000 / 142 * 200

weight of Na2SO4 = 142*200*0.2/1000

                                 =  5.68gm

 

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