How would you calculate the molarity of the following solution: 0.900 mol of $Na_2S$ in 1.10 L of solution?

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Answer 1 · 2016-08-30T07:34:27

$"Molarity"$ $=$ $"Moles of sodium sulfide"/"Volume of solution"$

$=$ $(0.900*mol)/(1.10*L)$ $=$ $0.818*mol*L^-1$ with respect to $"sodium sulfide"$ . What is the concentration with respect to $Na^+(aq)$ ion?

How would you calculate the molarity of the following solution: 0.900 mol of Na_2SNa_2S in 1.10 L of solution?