1.

How many gram of HCl will be present in 150 ml of its 0.52 M solution

Answer»

2.84 GM
5.70 gm
8.50 gm
3.65 gm

Solution :`M = (W)/(mxx V(L)), 0.52 = (w)/(36.5xx0.15), w = 2.84 gm`


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