Calculating number of moles of a given Ca(NO3)2:
"n(Ca(NO_3)_2) = \\frac{m(Ca(NO_3)_2)}{M(Ca(NO_3)_2)}=\\frac{12.5g}{(40 + 2\\cdot 14 + 6\\cdot 16)\\frac{g}{mol}}=\\frac{12.5g}{164\\frac{g}{mol}}=0.07622mol"
From equation for molar concentration "c=\\frac{n}{V}" the volume of calcium nitrate solution can be found:
"V(Ca(NO_3)_2)=\\frac{n(Ca(NO_3)_2)}{c(Ca(NO_3)_2)}=\\frac{0.07622mol}{0.250\\frac{mol}{L}}=0.305L=305mL"
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