A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

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A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

re volumetric flask. What is the molarity (M) of the resulting solution relative to the nitrate ion, NO3 1-

Chemistry

1 answer:

lorasvet [2.7K] · Answer 1 · 2026-03-22T23:29:17+03:00

Response:

For each nitrate-based salt, we assess the nitrate mole count, as depicted below: $M=0.213M$

$n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-$

$n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-$

$n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-$

Calcium nitrate contributes two moles of nitrate, sodium nitrate provides one, and aluminum nitrate contains three. Summing the moles gives us the total nitrate mole count:

$n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-$

Ultimately, we proceed to calculate the molarity:

$M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M$

Best,