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Calculate the amount of ammonium sulphate required to prepare 500ml of 1 molar solution of ammonium sulphate? (m.wt of $(NH_4)_2SO_4$ is 132.14gm/mol)

$\begin{array}{1 1}(a)\;66gm\\(b)\;132gm\\(c)\;198gm\\(d)\;212gm\end{array}$

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Answer: 66 g
Given Molarity of Ammonium Sulphate solution $ = 1$, Volume of solution is $500\;mL$ and the Molecular Weight $ = 132.14 \; g\;mol^{-1}$
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution, $M = \large\frac{\text{ Moles of solute }}{\text{Molarity Volume of solution in litre}}$
$\Rightarrow $ Molues of solute $= 1M \times 0.5\;L = 0.5\; mol$
$\Rightarrow 0.5\; mol = \large\frac{\text{Amount of Ammonium Sulphate}}{\text{Molar Mass of Ammonium Sulphate}}$
$\Rightarrow$ Amount of Ammonium Sulphate $= 0.5\;mol \times 131.14\; g\;mol^{-1} = 66\;g$
answered Feb 12, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai

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