1. A student prepares a 250.0 mL solution using 41.7 grams of potassium nitrite.
ID: 885154 • Letter: 1
Question
1. A student prepares a 250.0 mL solution using 41.7 grams of potassium nitrite. They then take 29.9 mL of this solution and dilute it to a final volume of 250.0 mL. How many grams of potassium nitrite are in a 36.2 mL sample of this final diluted solution?
2.
If you wanted to make a 4.50 M solution of aluminum nitrate using a 400.00 mL volumetric flask, how many grams of this salt would you need to use?
3.
H2SO4 + 2KOH 2H2O + K2SO4
How many mL of a 1.75 M solution of potassium hydroxide are required to titrate 15.72 mL of a 2.13 M solution of sulfuric acid?
If you wanted to make a 4.50 M solution of aluminum nitrate using a 400.00 mL volumetric flask, how many grams of this salt would you need to use?
3.
H2SO4 + 2KOH 2H2O + K2SO4
How many mL of a 1.75 M solution of potassium hydroxide are required to titrate 15.72 mL of a 2.13 M solution of sulfuric acid?
Explanation / Answer
Answer –
1) We are given, mass of KNO2 = 41.7 g , volume = 250.0 mL
First we need to calculate the moles of KNO2
Moles of KNO2 = 41.7 g / 85.104 g.mol-1
= 0.490 moles
So initial concentration of KNO2 = 0.490 moles / 0.250 L
= 1.96 M
So, M1= 1.96 M, V1 = 29.9 mL, V2 = 250.0 mL , M2 = ?
By dilution law
M1V1 = M2V2
So, M2 = M1V1/V2
= 1.96 M * 29.9 mL / 250.0 mL
= 0.234 M
So the final solution with 0.234 M and from this solution when we take 36.2 mL
So, moles of KNO2 = 0.234 M * 0.0362 L
= 0.008486 moles
So, mass of KNO2 = 0.008486 moles * 85.104 g.mol-1
= 0.722 g of KNO2
So, 0.722 grams of potassium nitrite are in a 36.2 mL sample of this final diluted solution.
2) Given, molarity of Al(NO3)3 = 4.50 M, volume = 400.0 mL
First we need to calculate the moles of Al(NO3)3
Moles of Al(NO3)3 = 4.50 M * 0.400 L
= 1.8 moles
So, mass of Al(NO3)3 = 1.8 mole * 212.996 g/mol
= 383.4 g of Al(NO3)3
3) Given, 15.72 mL of a 2.13 M solution of sulfuric acid, molarity of KOH = 1.75 M
Reaction - H2SO4 + 2KOH 2H2O + K2SO4
First we need to calculate the moles of H2SO4
Moles of H2SO4 = 2.13 M * 0.01572 L
= 0.0335 moles of H2SO4
From the above balanced equation
1 moles of H2SO4 = 2 moles of KOH
So, 0.0335 moles of H2SO4 = ?
= 0.0670 moles of KOH
So volume of KOH = 0.0670 moles / 1.75 M
= 0.0383 L
= 38.3 mL
So, 38.3 mL of a 1.75 M solution of potassium hydroxide are required to titrate 15.72 mL of a 2.13 M solution of sulfuric acid
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