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If sea water contains 63ppm $NO_3^{-1}$ and its density is 1.01g/ml,the approximate molarity of sea water is

$(a)\;0.001m\qquad(b)\;0.003m\qquad(c)\;0.002m\qquad(d)\;0.004m$

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$10^6g$ water or solution contains 63g $NO_3^-$=$\large\frac{63}{63}$mole$NO_3^-$(solution is dilute)
$\therefore 10^3g$ water contains=$\large\frac{1\times 10^3}{10^6}$
$\Rightarrow 0.001mole\;NO_3^-$
Weight of solution =weight of water=$10^3g$
$\therefore$ Volume of solution =$\large\frac{10^3}{1.01}$
$\Rightarrow 10^3ml$
$\Rightarrow 1litre$
Hence (a) is the correct answer.
answered Oct 28, 2013 by sreemathi.v
 

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