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The amount of $H_2SO_4$ present in 1200ml of 0.1m solution is


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$\large\frac{w}{m}$$\times 100=M\times V_{m2}$
$\large\frac{w}{98}$$\times 1000=1200\times 0.1$
$\therefore wH_2SO_4=11.76g$
Hence (a) is the correct answer.
answered Oct 28, 2013 by sreemathi.v

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