To 10ml of 1M BaCl2 solution 5ml of 0.5M H2SO4 is added ,BaSO4 is precipitated out.The amount of BaSO4 ppt will be ?
Dear Student
10ml of 1M BaCl2 solution + 5ml of 0.5M H2 SO4 -----> precipitated BaSO4
Calculate the amount of BaSO4 ?
Reaction will be:
(10ml) BaCl2 + (5ml) H2SO4 ------> (?) BaSO4 + 2HCl
1 mol + 1 mol --------> 1 mol + 2 mol
10 gm + 5 gm --------> ?
Because 1 mL = 1 gm, therefore, Mass of BaCl2 = 10 gm and Mass of H2SO4 = 5 gm
Molarity =
No. of Moles of BaCl2 =
=
[Molar Mass of BaCl 2 = 208.23 g/mol]
No. of Moles of BaCl2=
= 0.0480 mol
Molar Mass of BaSO4 = 233.38 g/mol
Molarity = = 0.0048 M
Therefore, 0.0048 =
Mass of BaSO4 = 0.0048 x 233.38
= 1.120 gm
Regards
10ml of 1M BaCl2 solution + 5ml of 0.5M H2 SO4 -----> precipitated BaSO4
Calculate the amount of BaSO4 ?
Reaction will be:
(10ml) BaCl2 + (5ml) H2SO4 ------> (?) BaSO4 + 2HCl
1 mol + 1 mol --------> 1 mol + 2 mol
10 gm + 5 gm --------> ?
Because 1 mL = 1 gm, therefore, Mass of BaCl2 = 10 gm and Mass of H2SO4 = 5 gm
Molarity =
No. of Moles of BaCl2 =
=
[Molar Mass of BaCl 2 = 208.23 g/mol]
No. of Moles of BaCl2=
= 0.0480 mol
Molar Mass of BaSO4 = 233.38 g/mol
Molarity = = 0.0048 M
Therefore, 0.0048 =
Mass of BaSO4 = 0.0048 x 233.38
= 1.120 gm
Regards