Question

a chemist prepares a solution of zinc nitrate (zn(no₃)₂) by measuring out 180. g of zinc nitrate into a 350. ml volumetric flask and filling the flask to the mark with water. calculate the concentration in mol/l of the chemists zinc nitrate solution. round your answer to 3 significant digits. mol/l

Explanation:

Step1: Calculate the molar mass of \(Zn(NO_3)_2\)

The molar mass of \(Zn\) is \(65.38\space g/mol\), \(N\) is \(14.01\space g/mol\), and \(O\) is \(16.00\space g/mol\).

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Step2: Calculate the number of moles of \(Zn(NO_3)_2\)

Use the formula \(n=\frac{m}{M}\), where \(m = 180.\space g\) and \(M=189.4\space g/mol\)

$$n=\frac{180.}{189.4}\space mol\approx0.950\space mol$$

Step3: Convert the volume of the solution to liters

Given \(V = 350.\space mL\), use the conversion \(1\space L=1000\space mL\), so \(V=\frac{350.}{1000}\space L = 0.350\space L\)

Step4: Calculate the molarity of the solution

Use the formula \(c=\frac{n}{V}\), where \(n = 0.950\space mol\) and \(V=0.350\space L\)

$$c=\frac{0.950}{0.350}\space mol/L\approx2.71\space mol/L$$

Answer:

\(2.71\space mol/L\)