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標題:
chemistrry QQQ
發問:
a) 0.50g of a sample of solid sodium hydroxide containing some sodium chloride is dissolved in water. 22.0cm3 of 0.50M nitric acid (HNO3) are required to complete neutralize the sample . Calculate the percentage purity of the sodium hydroxide in the sample. 更新: p.s. i don't understand why ''' 0.44/0.5 ''???????????? THZ 更新 2: 其實我有點不明白,,為何0.44(the mass of NaOH) 除0.5 (the mass of sodium hydroxide)就可得知% PURITY??
最佳解答:
No. of moles of HNO3 = 0.5 x 0.022 = 0.011 From the eqn NaOH + HNO3 → NaNO3 + H2O 1 mole of HNO3 completely neutralizes 1 mole of NaOH So no. of moles of NaOH = 0.011, with mass = 0.011 x 40 = 0.44 g Hence percentage purity = 0.44/0.5 x 100% = 88% 2011-11-30 15:35:19 補充： Becos the sample is 0.5 g in which it contains 0.44g being NaOH. 2011-12-01 00:20:07 補充： "0.50g of a sample of solid sodium hydroxide" 呢句意思係: 0.5g 係一個 sample, 但唔係 pure. 為求當中 NaOH 的實際量, 就用 titration 做個 result 出黎.
其他解答:
chemistrry QQQ
發問:
a) 0.50g of a sample of solid sodium hydroxide containing some sodium chloride is dissolved in water. 22.0cm3 of 0.50M nitric acid (HNO3) are required to complete neutralize the sample . Calculate the percentage purity of the sodium hydroxide in the sample. 更新: p.s. i don't understand why ''' 0.44/0.5 ''???????????? THZ 更新 2: 其實我有點不明白,,為何0.44(the mass of NaOH) 除0.5 (the mass of sodium hydroxide)就可得知% PURITY??
最佳解答:
No. of moles of HNO3 = 0.5 x 0.022 = 0.011 From the eqn NaOH + HNO3 → NaNO3 + H2O 1 mole of HNO3 completely neutralizes 1 mole of NaOH So no. of moles of NaOH = 0.011, with mass = 0.011 x 40 = 0.44 g Hence percentage purity = 0.44/0.5 x 100% = 88% 2011-11-30 15:35:19 補充： Becos the sample is 0.5 g in which it contains 0.44g being NaOH. 2011-12-01 00:20:07 補充： "0.50g of a sample of solid sodium hydroxide" 呢句意思係: 0.5g 係一個 sample, 但唔係 pure. 為求當中 NaOH 的實際量, 就用 titration 做個 result 出黎.
其他解答:
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