chemistry

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A laboratory procedure calls for making 400.0 mL of a 1.1 M NnO3 solution. What mass of NaNO3 (in g) is needed?

  • chemistry -

    M = moles/L
    Solve for moles.

    moles = grams/molar mass
    solve for grams.

  • chemistry -

    I dunno......I hate chemistry!

  • chemistry -

    please

  • chemistry -

    All you need to do is to substitute your problem numbers into the equations I've given you.
    M = moles/L
    1.1 = moles/0.400 L
    Solve for moles, the only unknown.
    moles = 1.1 x 0.400 = 0.440 mols.

    Second equation:
    moles = grams/molar mass
    moles from above = 0.440
    molar mass NaNO3 is about 85 (23+14+3*16) = about 85 but you need to confirm that.
    So 0.440 = grams/85
    Solve for grams, the only unknown.
    g = 0.440 x 85 =37.4 grams NaNO3 in 400 mL (0.400 L).

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