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The volume required to prepare 2.0L of 3.0M $H_2SO_4$ solution from 96% $H_2SO_4$ solution (density $1.83gmL^{-1}$) is

$\begin{array}{1 1}242.0mL\\334.7mL\\385.2mL\\450.1mL\end{array} $

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Answer : 334.7mL
Amount of HCl in 2.0L of 3.0M solution =$(2.0L)(3.0molL^{-1})=6.0mol$
Mass of $H_2SO_4$ in this solution=$(6.0mol)(98gmol^{-1})=588g$
Mass of 96% $H_2SO_4$ solution containing 588g of $H_2SO_4=588\times 100g/96=612.5g$
Volume of 96% $H_2SO_4$ solution weighing $612.5g/1.83gmL^{-1}=334.7mL$
answered Jul 22, 2014 by sreemathi.v

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