In order to prepare 100 cm³ of 0.250 M barium chloride solut... — Chemistry | Mockzy | Mockzy

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In order to prepare 100 cm³ of 0.250 M barium chloride solution the amount of BaCl₂·2H₂O required will be (Ba = 137) -

A

0.250 moles

B

0.0025 moles

C

2.5 moles

D

6.1 gram of BaCl₂·2H₂O

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