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5 votes
How much Sr(OH)2 • 8 H2O (M = 265.76) is needed to

prepare 250.0 mL of solution in which [OH–
] = 0.100 M?

User Sam Lu
by
8.3k points

1 Answer

3 votes
First, you need to find the number of moles of OH⁻ in a 250mL solution of 0.100M OH⁻. To do this, multiply 0.250L by 0.100M to get 0.025mol OH⁻. Then you use the fact that 1 mole of Sr(OH₂)·8H₂O contains 2 moles of OH⁻ which means that 0.0125mol of Sr(OH)₂·8H₂O contains 0.025mol OH⁻ (0.025/2=0.0125). Then to find the amount of Sr(OH)₂·8H₂O is needed you multiply its molar mass (265.76g/mol) by 0.0125mol to get 3.322g.
Therefore you need 3.322g of Sr(OH)₂·8H₂O.

I hope that helps. Let me know if anything is unclear.
User Cherokee
by
7.4k points

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