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In order to prepare 100cm^(3) of 0.250M ...

In order to prepare `100cm^(3)` of `0.250M` barium chloride solution, the amount of `BaCI_(2).2H_(2)O` required will be

A

0.250 moles

B

.0025 moles

C

2.5 moles

D

`6.1g` of `BaCI_(2).2H_(2)O`

Text Solution

Verified by Experts

The correct Answer is:
D
Amount of `BaCI_(2).2H_(2)O` required
`= (1"mole")/(1000) xx 100 xx 0.2` moles
`= 0.025 xx 224 = 6.1g`
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