Question

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

Answer 1

`M=0.213M`

Step-by-step explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

`n_(NO_3^-)=20.00gCa(NO_3)_2*(1molCa(NO_3)_2)/(164.088 gCa(NO_3)_2) *(2molNO_3^-)/(1molCa(NO_3)_2) =0.244molNO_3^-`

`n_(NO_3^-)=10.00gNaNO_3*(1molNaNO_3)/(84.9947 gNaNO_3) *(1molNO_3^-)/(1molNaNO_3) =0.118molNO_3^-`

`n_(NO_3^-)=50.00gAl(NO_3)_3*(1molAl(NO_3)_3)/(212.996gAl(NO_3)_3) *(3molNO_3^-)/(1molAl(NO_3)_3) =0.704molNO_3^-`

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

`n_(NO_3^-)^(Tot)=0.244+0.118+0.704=1.066molNO_3^-`

Finally, we compute the molarity:

`M=(1.066molNO_3^-)/(5.00L) \\\\M=0.213M`

Regards.