Question

One mole of any substance contains 6.022 × 10²³ atoms/molecules. Number of molecules of \[\ce{H2SO4}\] present in 100 mL of 0.02 M \[\ce{H2SO4}\] solution is ______.

Options

• 2.044 × 10²⁰ molecules

• 6.022 × 10²³ molecules

• 1 × 10²³ molecules

• 12.044 × 10²³ molecules

Answer 1

One mole of any substance contains 6.022 × 10²³ atoms/molecules. Number of molecules of \[\ce{H2SO4}\] present in 100 mL of 0.02 M \[\ce{H2SO4}\] solution is 2.044 × 10²⁰ molecules.

Explanation:

The molarity (M) is given by the formula:

M = `n/(V_((L)))`

On substituting the values in the above equation:

0.02 M = `(n xx 1000 L)/100`

n = 0.002 mol

The number of molecules can be calculated as, number of moles = `"Number of molecules"/"Avogadro's number"`  ......(2)

On substituting the values in the above equation:

0.002 mol = `"Number of molecules"/(6.022 xx 10^23)`

Number of molecules = 0.002 × 6.022 × 10²³ = 12.044 × 10²⁰