The mass percent of water in a hydrate of mncl2 is 36.41%, what is the formula of the hydrate?

The formula of the hydrate is MnCl2.4H2O

calculation

find the moles of MnCl2 and H2O

moles= % composition/ molar mass

% composition of MnCl2 = 100- 36.41=63.59 %

moles of MnCl2 = 63.59/ 125.84= 0.505 moles

moles of H2O= 36.41/18 = 2.022 moles

find the mole ratio by dividing each moles with smallest moles (0.505)

that is MnCl2 = 0.505/0.505=1

H2O = 2.022/0505 = 4

therefore the formula = MnCl2.4H2O

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