1gm of AgNO3 is dissolved in 50ml of h2o. it is titrated with 50ml od sol. agi is precipitated is filtered of.the exces of ki is then titrated with 1/10M KIO3 in the presence of acidic medium until all I->>ICl. it reqires 50ml of the 1/10M KIO3 sol. 20ml of same KI sol reqire 30ml of 1/10M KIO3. determine % od AgNO3 in the sample.

KIO3+KI+HCl>>ICl+KCl+H2O

You need to repost. This time, watch spelling, start sentences with capital letters, and use proper formula writing (I know what h2o means although it isn't correct) but I don't know what agi means and I don't know what od means.

To determine the percentage of AgNO3 in the sample, we need to calculate the amount of AgNO3 reacted during the titration.

First, let's look at the given information:

1) A mass of 1gm AgNO3 is dissolved in 50ml of water.
2) The solution is titrated with a solution of unknown concentration (let's call it "X") that contains KI.
3) The excess KI is titrated with a known concentration solution of KIO3 (1/10M) in the presence of acidic medium until all I- ions are converted to ICl.
4) It requires 50ml of the 1/10M KIO3 solution to react with the excess KI.

To find the concentration of unknown solution X, we can use the second titration:

20ml of the same KI solution requires 30ml of the 1/10M KIO3 solution.

We can write the balanced equation for the second titration as follows:

2KI + HCl + KIO3 → ICl + KCl + H2O

Now, let's calculate the number of moles of KIO3 consumed in each titration:

For the first titration:
Number of moles of KIO3 used = Molarity × Volume (in liters)
Number of moles of KIO3 used = (1/10) × (50/1000) = 0.005 moles

For the second titration:
Number of moles of KIO3 used = (1/10) × (30/1000) = 0.003 moles

Since the reaction between KIO3 and KI is 1:1 in stoichiometry, the number of moles of KI used in each titration is equal to the number of moles of KIO3 used.

Now, let's calculate the number of moles of AgNO3 in the sample:

Moles of AgNO3 used in the first titration = Moles of KI = 0.005 moles
Moles of AgNO3 used in the second titration = Moles of KI = 0.003 moles

Total moles of AgNO3 used = 0.005 + 0.003 = 0.008 moles

Now, we can calculate the percentage of AgNO3 in the sample:

Percentage of AgNO3 = (Total moles of AgNO3 used / Moles of AgNO3 in the sample) × 100

Assuming the mass of AgNO3 is equal to the molar mass (g/mole) of AgNO3, we can calculate the mass of AgNO3 in the sample:

Mass of AgNO3 in the sample = Moles of AgNO3 in the sample × Molar mass of AgNO3

To calculate the percentage, we can use the formula:

Percentage of AgNO3 = (Mass of AgNO3 in the sample / Mass of the sample) × 100

Note: The mass of the sample can be calculated by the formula:

Mass of the sample = Mass of AgNO3 + Mass of water

Please provide the molar mass of AgNO3 to complete the calculation.