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✒️How many grams of NaBrO3 are required to prepare 250 ml of 0.2 N solution,when the product being NaBr?
‼️ Explanation is needed ‼️
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Answered by
2
Answer:
Molecular mass of NaBrO
3
=23+80+(3×16)=151
Each bromate ion takes-up 6 electrons; therefore,
Eq. mass of NaBrO
3
=
6
Mol.mass
=
6
151
Amount of NaBrO
3
in 85.5 mL 0.672 N solution
=
1000
0.672
×
6
15!
×85.5=1.446 g
Molarity=
n
Normality
=
6
0.672
=0.112 M
(ii) Each bromate ion takes-up 5 electrons; therefore,
Eq. mass of NaBrO
3
=
5
Mol.mass
=
5
151
Amount of NaBrO
3
in 85.5 mL 0.672 N solution
=
2
151
×
1000
0.672
×85.5
=1.7352 g
Molarity=
n
Normality
=
4
0.672
−0.1344 M.
Explanation:
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Explanation:
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