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A solution is prepared by dissolving 150.0g od magnesium chloride in 2.0L of water. What is the osmolarity of the resulting solution?

I need help showing how to get the anwser

2 Answers

  • mRNA
    Lv 5
    1 decade ago
    Favorite Answer

    For MgCl2 osmolarity, xc = v(MgCl2)[MgCl2]

    where v(MgCl2) = 3 (as MgCl2 dissociates into 3 species: Mg2+, 2 x Cl-)

    [MgCl2] = m(MgCl2) / V(soln)M(MgCl2)

    = (150.0) / {(2.0)(95.211)}

    = 0.7877 mol Lˉ¹

    ==> xc = (3)(0.7877)

    = 2.36 osmol Lˉ¹

    NB: This is an approximation as strictly xc = φ(X)n(X)[X], where φ is the osmotic coefficent and:

    φ = -n(H2O) ln{a(H2O)} / v(X)n(X)

    where n is moles, a is activity of water

  • 1 decade ago

    you need to work out the No. of moles of MgCl. This is done by mass/molecular mass. Then that number is divided by the volume in L. as molarity is the No. moles per L.

    I think it is better for you to do the number crunching by yourself. Or else your not really learning, just stealing my knowlege lol.

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