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6 May, 18:43

If you want to prepare 5 liters of a 0.35m solution of nh4cl, how many grams of salt

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  1. 6 May, 19:06
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    Answer is: mass fo ammonium chloride is 93.625 grams.

    V (NH₄Cl) = 5 L.

    c (NH₄Cl) = 0.35 M.

    n (NH₄Cl) = V (NH₄Cl) · c (NH₄Cl).

    n (NH₄Cl) = 5 L · 0.35 mol/L.

    n (NH₄Cl) = 1.75 mol.

    M (NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.

    m (NH₄Cl) = n (NH₄Cl) · M (NH₄Cl).

    m (NH₄Cl) = 1.75 mol · 53.5 g/mol.

    m (NH₄Cl) = 93.625 g.
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