% purity question

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username1398750
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#1
Report Thread starter 7 years ago
#1
'A solution was made by dissolving 7.5g of NaOH, containing an inert purity, in water and making up 250cm³ of solution. If 20cm³ of the solution is neutralised by 13cm³ of 1moldm-³ HNO3, calculate the % purity of the NaOH' :confused::confused::confused:
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#2
Report 7 years ago
#2
Calculate moles of HNO3 in the 13cm3 of 1moldm-3
Write the equation for the reaction of NaOH and HNO3 to get the reacting ratio. This lets you calculate the moles of NaOH that reacted. Convert those moles to grams of NaOH.
Calculate that mass as a percentage of 7.5 g
Post your answer here.
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#3
Report Thread starter 7 years ago
#3
93.06%?
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#4
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#4
(Original post by Isla-H)
93.06%?
No.
Sorry I forgot to put in the scale up.
Calculate moles of HNO3 in the 13cm3 of 1moldm-3
Write the equation for the reaction of NaOH and HNO3 to get the reacting ratio. This lets you calculate the moles of NaOH that reacted in the 20cm3.
Multiply this number by 250/20 to scale up to the number of moles in the original 250cm3
Convert those moles to grams of NaOH.

Calculate that mass as a percentage of 7.5 g

Put your working out here.

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#5
Report Thread starter 7 years ago
#5
So I convert 13cm³ into dm³, so the moles are 0.013mol.
The equation is NaOH + HNO3 =NaNO3 + H2O, so the ratio is just 1:1 and so NaOH also has 0.013mol.
0.25/0.02=12.5*0.013=0.1625
Mass=0.1625*40 (Mr?)=6.5g
(6.5/7.5) *100=86.66%?
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#6
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#6
(Original post by Isla-H)
So I convert 13cm³ into dm³, so the moles are 0.013mol.
The equation is NaOH + HNO3 =NaNO3 + H2O, so the ratio is just 1:1 and so NaOH also has 0.013mol.
0.25/0.02=12.5*0.013=0.1625
Mass=0.1625*40 (Mr?)=6.5g
(6.5/7.5) *100=86.66%?
Correct!
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#7
Report Thread starter 7 years ago
#7
Yay , thanks for your help!
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