On a shelf sits a bottle of NaCl solutions with a molar concentration of 2.50 M and a total volume of 300 mL. If 15.0 mL of the solution is added to 985 mL of pure water, what is the molar concentration of NaCl in the new solution

Given :

On a shelf sits a bottle of NaCl solutions with a molar concentration of 2.50 M and a total volume of 300 mL. If 15.0 mL of the solution is added to 985 mL of pure water .

To Find :

The molar concentration of NaCl in the new solution.

Solution :

Molarity M is given by :

Now , 1000 ml of NaCl contains 2.5 moles of NaCl .

So , 15 ml of NaCl contains :

New , volume V = 15 + 985 = 1000 ml = 1 L .

So , putting value of n and V in above equation :

Therefore ,  the molar concentration of NaCl in the new solution is 0.0375 M .

Hence , this is the required solution .

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