$(a)\;0.2m\qquad(b)\;0.7m\qquad(c)\;0.8m\qquad(d)\;1.2m$

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One mole of $CuSO_4$ gives 1mole of $CU^{2+}$ and 1 mole of $SO_4^{2-}$.Also 1 mole of $Al_2(SO_4)_3$ gives 2 mole of $Al^{3+}$ and 3mole of $SO_4^{2-}$ .Thus total moles of all the ions present in solution having 0.1m of $CUSO_4$ and 0.1M of $Al_2(SO_4)_3$ is

$0.1+0.1+0.2+0.3=0.7$

Hence (b) is the correct option.

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