Question

CALCULATE RESULTING MOLARITY OF A SOLUTION:
1)
1500ML 1M HCL + 18.25gm Hcl
2) 200ml 1M HCl + 300ml h20
3) 200ml 1M HCl + 100ml 0.5M h2so4

Answer 1

Answer:

1500+18.25=33.25 200+300+20=520 200+100=300 +0.530.5 2+4=6

Answer 2

Answer:

1). 1.34  2). 0.4   3). 0.8167

Explanation:

1). no. of moles in 1500 ml 1M HCl = 1500*1 = 1500 milimoles

   no. of moles in 18.25g HCl = given weight of HCl/ =molar weight of HCl

                                            => 18.25g/36.5g= 1/2 mole = 500 milimoles

   Molarity(M) = total no. of moles/volume of solution

                  => (1500+500)/1500 = 4/3 = 1.34

2).  no. of moles in 200ml 1M HCl = 200*1 = 200 milimoles

         Molarity(M) = no. of moles/volume of solution

                           => 200/(200+300)= 200/500=2/5 = 0.4

3).  Here concentration of H+ would be considered

     =>no. of moles of H+ in 200ml 1M HCl = 200*1 = 200 milimoles

     =>no. of moles of H+ in 100ml 0.5M H2SO4 = 0.5*100 = 50 milimole

Molarity(M) = no. of moles/Volume of solution

                  => (200+50)/(200+100)= 250/300 = 5/6 = 0.8167