A13.97 g sample of nabr contains 22.34% na by mass. considering the law of constant composition (definite proportions) how many grams of sodium does a 5.75 g sample of sodium bromide contain?

As per the law of constant composition, a given sample will always contain the same number of elements that combine in the same mass proportion.

Therefore if a sample of 13.97 g of NaBr contains 22.39 % of Na by mass then,  a sample of 5.75 g of NaBr would also contain 22.39% Na by mass

Hence:

Mass of Na = 5.75 g * 22.39/100 = 1.287 g

5.75 g of NaBr would contain 1.29 g of Na

i'm assuming that the correct answer is indeed the first one/ "a"..

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