What mass of solute is required to prepare a $0.875"molal"$ solution from $534mL$ of water solvent?

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Answer 1 · 2016-08-12T18:35:57

Approx. $27*g$ . I assume you mean a $"molal concentration"$ .

$"Molality"="moles of solute"/"kg of solvent"$

Thus $0.875*mol*kg^-1$ $=$ $"number of moles of NaCl"/(0.534*kg)$

And $"number of moles of NaCl"$ $=$ $0.875*mol*kg^-1xx0.534*kg$ $=$ $0.467*mol$

$"Grams of NaCl"=0.467*molxx58.44*g*mol^-1$ . The $"molar concentration"$ would be to all intents and purposes identical.

What mass of solute is required to prepare a 0.875*"molal"0.875*"molal" solution from 534*mL534*mL of water solvent?