the experimental yield of a compound is 30 g and the true yield is 40 g what is the percentage error in calculating the yield

25%.

Explanation:

The following data were obtained from the question:

Experimental yield = 30 g

Actual yield = 40 g

Percentage error =..?

Thus, we can calculate the percentage error as follow:

Percentage error = |Experimental – Actual | / Actual × 100

Percentage error = |30 – 40|/40 × 100

Percentage error = 10/40 × 100

Percentage error = 25 %

Therefore, the percentage error in calculating the yield is 25%

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole

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