The molecular weight of calcium carbonate, \text{CaCO}_3CaCO 3 start text, C, a, C, O, end text, start subscript, 3, end subscript, is 100.09\,\dfrac{\text g}{\text{mol}}100.09 mol g 100, point, 09, start fraction, start text, g, end text, divided by, start text, m, o, l, end text, end fraction. How many moles of calcium carbonate are in 63.8\,\text g63.8g63, point, 8, start text, g, end text of calcium carbonate?

The number of moles of calcium carbonate is 0.637 moles

Explanation:

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

We are given:

Given mass of calcium carbonate = 63.8 g

Molar mass of calcium carbonate = 100.09 g/mol

Putting values in above equation, we get:

Hence, the number of moles of calcium carbonate is 0.637 moles

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