Question

Calculate the number of molecules present in 11.5dm3 of nitrogen at stp.

Answer 1

Answer:

8.603x1023

Explanation:

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023 1 litres of N2 at S.T.P. contains molecules = (6.022 x 1023)/22.4

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023 1 litres of N2 at S.T.P. contains molecules = (6.022 x 1023)/22.4

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023 1 litres of N2 at S.T.P. contains molecules = (6.022 x 1023)/22.4 32 litres of N2 at S.T.P. contains molecules = (32 x 6.022 x 1023)/22.4

We know that one mole of N2 at S.T.P. occpies 22.4 litres and contains 6.022 x 1023 molecules of N2.Therefore, 22.4 litres of N2 at S.T.P. contains molecules = 6.022 x 1023 1 litres of N2 at S.T.P. contains molecules = (6.022 x 1023)/22.4 32 litres of N2 at S.T.P. contains molecules = (32 x 6.022 x 1023)/22.4 = 8.603 x1023