Using equation for molar concentration "c=\\frac{n}{V}"
, where c - molar concentration, n - amount of compound in moles, V - volume of solution
we can determine volume of solution:
"V(CuNO_3)=\\frac{n(CuNO_3)}{c(CuNO_3)}"
The amount n(CuNO3) can be found from the equation
"n(CuNO_3)=\\frac{m(CuNO_3)}{M(CuNO_3)}"
,where m - mass of CuNO3 and M(CuNO3) is molar mass of CuNO3.
Thus "V(CuNO_3)=\\frac{n(CuNO_3)}{c(CuNO_3)}=\\frac{m(CuNO_3)}{M(CuNO_3) \\cdot c(CuNO_3)}=\\frac{3.58g}{126\\frac{g}{mol} \\cdot 0.670 \\frac{mol}{L}}=0.042L=42mL"
By the way, copper (I) nitrate CuNO3 doesn't exist in pure form without ligands, thus the problem in the given wording doesn't make any practical sense.
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