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19 March, 19:01

Potassium fluoride (kf), a salt, has a molecular weight of 58.10 grams. how many grams would be needed to mix 1.0 liter of 0.10 m salt solution? 0.17 g 580 g 17 g 5.8 g

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  1. 19 March, 19:48
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    N (KF) = m (KF) / M (KF)

    n (KF) = vc

    m (KF) / M (KF) = vc

    m (KF) = M (KF) vc

    m (KF) = 58.10*1.0*0.10=5.81 g≈ 5.8 g
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