Question

number of iodine atoms present in 50 ml of 0.1 M KL solution is​

Answer 1

Answer:

1. Molarity = number of moles ×1000400.1 M

=number of moles ×1004number of moles

=0.1 × 4100=

0.004 moles2.

Number of I2 molecules = 6.023 × 1023 ×0.004 =2.41×10213.

Number of I atoms = 2×2.41×1021 = 4.82 × 1021

Hope it helps you.

Answer 2

Number of iodine atoms present in given amount of KI solution is $3.011\times 10^{21}$

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KI solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of KI}\times 1000}{50}\\\\\text{Moles of KI}=\frac{0.1\times 50}{1000}=0.005mol$

1 mole of KI is made from 1 mole of potassium atom and 1 mole of iodine atom

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 0.005 moles of KI will contain $(1\times 0.005\times 6.022\times 10^{23})=3.011\times 10^{21}$ number of iodine atoms

Number of iodine atoms = $3.011\times 10^{21}$

Learn more about molarity and mole concept:

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