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How many gram of HCl will be present in 150 ml of its 0.52 M solutionA. 2.84 gmB. 5.70 gmC. 8.50 gmD. 3.65 gm

Answer» Correct Answer - A
`M = (w)/(mxx V(l)), 0.52 = (w)/(36.5xx0.15), w = 2.84 gm`


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