First, let's write down the given data for clarity:

• Mass of commercial AgNO₃ = 1 g


• Volume of the KI solution used = 50 ml


• Volume of (M/10) KIO₃ solution used to titrate excess KI = 50 ml


• Volume of (M/10) KIO₃ solution used to titrate 20 ml of KI solution = 30 ml

From the reaction given:

$$ KIO_3 + 2KI + 6HCl \to 3ICl + 3KCl + 3H_2O $$

It can be seen that 1 mole of KIO₃ reacts with 2 moles of KI. The molarity of KIO₃ solution is also given as (M/10), which means 0.1 M.

First, let's calculate the moles of KIO₃ used to titrate the excess KI present after the precipitation of AgI:

$$ \text{Moles of KIO}_3 = \text{Volume} \times \text{Molarity} $$

$$ = 50 \text{ ml} \times 0.1 \text{ M} $$

Since 1 liter is 1000 ml, we convert ml to liters:

$$ = \frac{50}{1000} \text{ L} \times \text{0.1 M} $$

$$ = 0.005 \text{ moles} $$

Now, 1 mole of KIO₃ reacts with 2 moles of KI; therefore, 0.005 moles of KIO₃ will react with:

$$ 0.005 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $$

$$ = 0.01 \text{ moles KI} $$

So, the excess amount of KI after the precipitation of AgI is 0.01 moles.

Next, we need to find out the amount of KI present in the 20 ml of KI solution that reacts with 30 ml of the KIO₃ solution.

$$ \text{Moles of KIO}_3 (for 20 \text{ ml KI solution}) = \text{Volume} \times \text{Molarity} $$

$$ = 30 \text{ ml} \times 0.1 \text{ M} $$

$$ = \frac{30}{1000} \text{ L} \times \text{0.1 M} $$

$$ = 0.003 \text{ moles KIO}_3 $$

This will react with twice the amount of KI:

$$ 0.003 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $$

$$ = 0.006 \text{ moles KI} $$

This amount is present in 20 ml of the KI solution. To find the amount of KI in the initial 50 ml used for the reaction with AgNO₃, we set up a proportion, assuming the KI solution is uniform in concentration:

$$ \frac{0.006 \text{ moles KI}}{20 \text{ ml}} = \frac{x \text{ moles KI}}{50 \text{ ml}} $$

Solving for x:

$$ x = \frac{0.006 \text{ moles KI} \times 50 \text{ ml}}{20 \text{ ml}} $$

$$ x = 0.015 \text{ moles KI} $$

From the reaction between AgNO₃ and KI:

$$ AgNO_3 + KI \to AgI \downarrow + KNO_3 $$

You can see that 1 mole of AgNO₃ reacts with 1 mole of KI. If 0.01 moles of KI remained after the reaction, the amount of KI that reacted with AgNO₃ is:

$$ 0.015 \text{ moles KI (total)} - 0.01 \text{ moles KI (excess)} $$

$$ = 0.005 \text{ moles KI reacted with AgNO}_3 $$

Therefore, the moles of AgNO₃ in the 1 gram commercial sample is also 0.005 moles, since the molar ratio of AgNO₃ to KI is 1:1. Now, let's calculate the mass of 0.005 moles of pure AgNO3:

The molar mass of AgNO₃ = Atomic mass of Ag + Atomic mass of N + 3 x Atomic mass of O

= 108 + 14 + 3 x 16 = 170 g/mol

The mass of 0.005 moles of AgNO₃ is:

$$ 0.005 \text{ moles} \times 170 \text{ g/mol} $$

$$ = 0.85 \text{ grams} $$

Finally, to find the percentage of AgNO₃ in the sample, we divide the mass of pure AgNO₃ by the mass of the commercial sample and multiply by 100%.

$$ \text{Percentage of AgNO}_3 = \frac{0.8493655 \text{ g}}{1 \text{ g}} \times 100\% $$

$$ = 85\% $$

Thus, the sample contains approximately 85 AgNO₃.