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Q)

What will be the molarity of a solution , which contains 5.85 g of Nacl per 500 mL?

$\begin{array}{1 1} (a)0.01\;moles L^{-1}\\ (b)0.2\;mole\;L^{-1}\\ (c) 0.07\;moles \;L^{-1}\\ (d) 0.08\;moles\;L^{-1} \end{array} $

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A)
Solution :
$M = \large\frac{\text{no of moles of solute}}{\text{volume of solution in L}}$
$\qquad= \large\frac{5.85/5.85}{0.5\;L}$
$\qquad= \large\frac{0.1}{0.5}$
$\qquad= 0.2\;mol\;L^{-1}$
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