Asolution is made by mixing of acetyl bromide and of heptane . calculate the mole fraction of acetyl bromide in this solution. be sure your answer has the correct number of significant digits.

The question is incomplete, here is the complete question:

A solution is made by mixing 127.g of acetyl bromide and 90.g of heptane . Calculate the mole fraction of acetyl bromide in this solution. Round your answer to 3 significant digits.

The mole fraction of acetyl bromide is 0.534 and heptane is 0.466

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of acetyl bromide = 127. g

Molar mass of acetyl bromide = 123 g/mol

Putting values in equation 1, we get:

Given mass of heptane = 90. g

Molar mass of heptane = 100.2 g/mol

Putting values in equation 1, we get:

Mole fraction of a substance is given by:

For acetyl bromide:

For heptane:

Hence, the mole fraction of acetyl bromide is 0.534 and heptane is 0.466

the answer is c. nacl

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