3. 3. A 90.-mL sample of juice was titrated with the I2(aq) solution described above using a sb buret. The initial reading of the sb buret was 58.3 mL. When the endpoint was reached, the reading on the sb buret was 26.9 mL. How many mg of Vitamin C were in the juice sample

2765mg of Vitamin C assuming the concentration of I2 = 0.500M

Explanation:

Iodine, I2, reacts with vitamin C in a ratio of 1:1 (The moles of  I2 added = Moles of Vitamin C in the sample). To solve this question we must find the difference in volume from the initial reading to the endpoint reading. Now, ASSUMING the molar concentration of the I2(aq) solution is 0.500M we can find the moles of I2 = Moles of Vitamin C. With its molar mass -176.12g/mol- we can find its mass:

Moles I2:

Volume  = 58.3mL - 26.9mL = 31.4mL = 0.0314L * (0.500mol / L) = 0.0157 moles I2 = Moles Vitamin C

Mass Vitamin C:

0.0157 moles * (176.12g/mol) = 2.765g of Vitamin C * (1000mg/1g) =

2765mg of Vitamin C assuming the concentration of I2 = 0.500M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

it is b that is the answer.