It takes 42.25 ml of 0.0500 M Na2S2O3 solution to completely react with the iodine present in a 150.0 ml iodine solution. How many grams of iodine (I2) were initially present in the sample

There were 0.268 grams I2 present

Explanation:

Step 1: Data given

Volume of Na2SO3 = 42.25 mL = 0.04225 L

Molarity of Na2SO3 = 0.0500 M

Volume of iodine = 150.0 mL

Step 2: The balanced equation

I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI

Step 3: Calculate moles of Na2SO3

Moles Na2SO3 = molarity * volume

Moles Na2SO3 = 0.0500 M * 0.04225 L

Moles Na2SO3 = 0.0021125 moles

Step 4: Calculate moles I2

For 1 mol I2 we need 2 moles Na2SO3

For 0.0021125 moles Na2SO3 we have 0.00105625

Step 5: Calculate mass of I2

Mass I2 = moles * molar mass

Mass I2 = 0.00105625 * 253.8 g/mol

Mass I2 = 0.268 grams

There were 0.268 grams I2 present

i'm assuming that the correct answer is indeed the first one/ "a"..