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? asked in Science & MathematicsChemistry · 1 decade ago

What volume (mLs) of a 0.29 M solution of sodium bromide would contain 3.8 g of sodium bromide?

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  • Raina
    Lv 5
    1 decade ago
    Favorite Answer

    NaBr = 102.9 g/mol

    3.8 g / 102.9g/mol = 0.0369 mol NaBr

    0.29 = 0.0369 / L

    0.29(L) = 0.0369

    L = 0.127 L

    0.127 x 1000 = 127 mL

    Source(s): Chemistry Student
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  • 1 decade ago

    NaBr has a molar mass of 103 g/mol.

    3.8 g NaBr x (1 mol / 103 g) = 0.0369 mol

    M = mol / L

    L = mol / M = 0.0369 mol / 0.29 mol/L = 0.127 L = 127 mL

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