What HCl concentration (in mol/L) is obtained when 8 ( Solution Final concentrat

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Final concentration(Mf) = n/V => Mf = ((initial concentration of HCL)*(volume of that solution taken))/(total volume) => Mf = (Mi)*(Vi)/V We have, Mi = 11 M, Vi = 8 ml and V = 250 ml => Mf = (0.011)*(8)/(250) => Mf = 0.352 M Now, Mf = (Mi)*(Vi)/V => ln(Mf) = ln(Mi) + ln(Vi) - ln(V) differentiating both sides, (for error calculation, all the errors are to be added up) d(Mf)/Mf = d(Mi)/Mi + d(Vi)/Vi + dV/V We have, d(Mi) = 0.75 % of 11 = 0.0825 M d(Vi) = 0.05 ml d(V) = 0.5 ml => d(Mf)/0.352 = (0.0825/11) + (0.05/8) + (0.5/250) => d(Mf) = 0.005544 M = 0.006 M => final concentration = (0.352 ± 0.006) M

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