What is the mass percent of sodium fluoride solution prepared by dissolving 0.64 moles of sodium fluoride into 63.5 grams of water?

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What is the mass percent of sodium fluoride solution prepared by dissolving 0.64 moles of sodium fluoride into 63.5 grams of water?

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Answer 1 · 2018-04-07T18:00:43

$m a s s % = 29.7 %$ $mass %=29.7%$

Explanation:

moles of sodium fluoride = $\frac{W (g i v e n m a s s)}{M (m o l a r m a s s)}$ $(W(given mass))/(M(molar mass))$
$0.64 = \frac{W}{42}$ $0.64=W/42$
$W = 26.88$ $W=26.88$ gm
$m a s s % = \frac{m a s s_{s o l u t e}}{m a s s_{s o l u t i o n}} \cdot 100 = \frac{26.88}{90.48} \cdot 100 = 29.7 %$ $mass %=[mass_(solute)]/[mass_(solution)]*100=26.88/90.48*100=29.7%$

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What is the mass percent of sodium fluoride solution prepared by dissolving 0.64 moles of sodium fluoride into 63.5 grams of water?

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