In 500 mL of 0.2 M NaCl solution,1500 mL | vedclass

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(NONE) Total volume $= 500 \ mL + 1500 \ mL = 2000 \ mL = 2 \ L$. Moles of $Na^{+}$ ions $= 0.2 \ M 	imes 0.5 \ L = 0.1 \ mol$. Concentration of $Na^

Vedclass · Accepted Answer

$[Na^{+}] = 0.05 \ M$

Vedclass · Answer

(NONE) Total volume $= 500 \ mL + 1500 \ mL = 2000 \ mL = 2 \ L$. Moles of $Na^{+}$ ions $= 0.2 \ M 	imes 0.5 \ L = 0.1 \ mol$. Concentration of $Na^{+}$ ions $= \frac{0.1 \ mol}{2 \ L} = 0.05 \ M$. Moles of $Mg^{2+}$ ions $= 0.4 \ M 	imes 1.5 \ L = 0.6 \ mol$. Concentration of $Mg^{2+}$ ions $= \frac{0.6 \ mol}{2 \ L} = 0.3 \ M$. Mass concentration of $Mg^{2+}$ $= 0.3 \ mol/L 	imes 24 \ g/mol = 7.2 \ g/L$. Moles of $Cl^{-}$ ions $= (0.2 	imes 0.5) + (2 	imes 0.4 	imes 1.5) = 0.1 + 1.2 = 1.3 \ mol$. Concentration of $Cl^{-}$ ions $= \frac{1.3 \ mol}{2 \ L} = 0.65 \ M$. Since all calculated values match the options,there is no incorrect statement provided in the options.

In $500 \ mL$ of $0.2 \ M \ NaCl$ solution,$1500 \ mL$ of $0.4 \ M \ MgCl_2$ solution is mixed. The only incorrect final concentration of ion is (assume volume to be additive)

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