MediumChemistryConcentration Terms
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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
Explanation
Moles required = 0.25 × 0.1 = 0.025 mol. Molar mass BaCl₂·2H₂O = 244 g mol⁻¹. Mass = 0.025×244 = 6.1 g.
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