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0.3g of an oxalate salt was dissolved in 100 ml solution. The solution required 90 ml of N/20 KMnO4 for complete oxidation. The % of oxalateion in salt is:
- a)33%
- b)66%
- c)70%
- d)40%
Correct answer is option 'B'. Can you explain this answer?
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0.3g of an oxalate salt was dissolved in 100 ml solution. The solution...
To find the percentage of oxalate ion in the given salt, we need to determine the number of moles of oxalate ion and the number of moles of the entire salt.
First, let's determine the number of moles of KMnO4 used in the titration. Since the solution required 90 ml of N/20 KMnO4, we can calculate the number of moles of KMnO4 using the formula:
Moles of KMnO4 = Volume of KMnO4 (in liters) x Normality of KMnO4
Given that the volume of KMnO4 used is 90 ml = 0.09 L and the normality of KMnO4 is N/20 = 1/20 = 0.05 N, we can substitute these values into the formula:
Moles of KMnO4 = 0.09 L x 0.05 N = 0.0045 moles
Since the reaction between KMnO4 and oxalate ion is 1:5, we can determine the number of moles of oxalate ion in the titrated solution as follows:
Moles of oxalate ion = 5 x Moles of KMnO4 = 5 x 0.0045 = 0.0225 moles
Now, let's determine the number of moles of the salt. We know that 0.3 g of the salt is dissolved in 100 ml of solution. We can calculate the molar mass of the salt using the periodic table and convert the grams to moles:
Molar mass of the salt = (Molar mass of K + Molar mass of Mn + 4 x Molar mass of O) = (39.1 + 54.9 + 4 x 16) = 39.1 + 54.9 + 64 = 158 g/mol
Moles of the salt = Mass of the salt (in grams) / Molar mass of the salt
Given that the mass of the salt is 0.3 g, we can substitute these values into the formula:
Moles of the salt = 0.3 g / 158 g/mol ≈ 0.0019 moles
Finally, we can calculate the percentage of oxalate ion in the salt using the formula:
% Oxalate ion = (Moles of oxalate ion / Moles of the salt) x 100
Substituting the values we calculated earlier:
% Oxalate ion = (0.0225 moles / 0.0019 moles) x 100 ≈ 1184.21%
However, the maximum percentage can only be 100%, so we can conclude that the given options are incorrect.
Therefore, the correct answer cannot be determined using the information provided in the question, and none of the options (A, B, C, D) are correct.
First, let's determine the number of moles of KMnO4 used in the titration. Since the solution required 90 ml of N/20 KMnO4, we can calculate the number of moles of KMnO4 using the formula:
Moles of KMnO4 = Volume of KMnO4 (in liters) x Normality of KMnO4
Given that the volume of KMnO4 used is 90 ml = 0.09 L and the normality of KMnO4 is N/20 = 1/20 = 0.05 N, we can substitute these values into the formula:
Moles of KMnO4 = 0.09 L x 0.05 N = 0.0045 moles
Since the reaction between KMnO4 and oxalate ion is 1:5, we can determine the number of moles of oxalate ion in the titrated solution as follows:
Moles of oxalate ion = 5 x Moles of KMnO4 = 5 x 0.0045 = 0.0225 moles
Now, let's determine the number of moles of the salt. We know that 0.3 g of the salt is dissolved in 100 ml of solution. We can calculate the molar mass of the salt using the periodic table and convert the grams to moles:
Molar mass of the salt = (Molar mass of K + Molar mass of Mn + 4 x Molar mass of O) = (39.1 + 54.9 + 4 x 16) = 39.1 + 54.9 + 64 = 158 g/mol
Moles of the salt = Mass of the salt (in grams) / Molar mass of the salt
Given that the mass of the salt is 0.3 g, we can substitute these values into the formula:
Moles of the salt = 0.3 g / 158 g/mol ≈ 0.0019 moles
Finally, we can calculate the percentage of oxalate ion in the salt using the formula:
% Oxalate ion = (Moles of oxalate ion / Moles of the salt) x 100
Substituting the values we calculated earlier:
% Oxalate ion = (0.0225 moles / 0.0019 moles) x 100 ≈ 1184.21%
However, the maximum percentage can only be 100%, so we can conclude that the given options are incorrect.
Therefore, the correct answer cannot be determined using the information provided in the question, and none of the options (A, B, C, D) are correct.
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0.3g of an oxalate salt was dissolved in 100 ml solution. The solution...
No eq. kmno4=90*1/2=4.5, no eq oxalate salt=.3 , so percentage= 0.3/4.5=66%
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0.3g of an oxalate salt was dissolved in 100 ml solution. The solution required 90 ml of N/20 KMnO4 for complete oxidation. The % of oxalateion in salt is:a)33%b)66%c)70%d)40%Correct answer is option 'B'. Can you explain this answer?
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