Question

12. a. How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP?

b. 4 litres of water are added to 2L of 6 molar HCl solutions. What is the molarity of resulting solution?

c. What volume of 10M HCL and 3M HCI should be mixed to obtain 1L of 6M HCL solution?​

Answer 1

A:The equation is : 2KClO

3

→2KCl+3O

2

2moles→3moles

i.e. 2(39+35.5+48)→3×22.4 of oxygen

245g of KClO

3

3×22.4lit

i.e. 3×22.4 lit of O

2

→245g of KClO

3

∴2.24 lit of O

2

→

3×22.4

245

×2.24

=8.17g

B:Total volume of the solution after mixing =(4+2)=6L

By applying molarity equation :

M1V1(dilute)≡M2V2(conc.)

M1×6=6×2orM1=6×26=2M.

C:Applying molarity equation:

Therefore Vol. of 10 M HCl required = 0.428 L = 428 mL. Vol. of 3 M HCl required = 1-0.428 = 0.527 L = 572 mLRead more on Sarthaks.com - https://www.sarthaks.com/64253/what-volume-of-10-m-hcl-and-3-m-hcl-should-be-mixed-to-get-1l-of-6-m-hcl-solution