How many grams of barium chloride (BaCl_(2)) are needed to prepare 100 – InterviewSolution

1.	How many grams of barium chloride (BaCl_(2)) are needed to prepare 100cm^(3) of 0.250 mL of a 0.50 M BaCl_(2) solution?
Answer» SOLUTION :Molecular mass of `BaCl_(2)=137+71=208u`. For 1000 `CM^(3)` of 1 M `BaCl_(2)` so. Mass of `BaCl_(2)` needed `=(208)/(1000)x100xx0.25g=5.20g`

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