Question

questions and problems
q5 a 15.0 - ml sample of nacl solution has a mass of 15.78 g. after the nacl solu - tion is evaporated to dryness, the dry salt residue has a mass of 3.26 g. calculatethe following concentrations for the nacl solution.
a. % (m/m)
b. % (m/v)
c. molarity (m)
q6 how many grams of ki are in 25.0 ml of a 3.0% (m/v) ki solution?
q7 how many milliliters of a 2.5 m mgcl₂ solution contain 17.5 g mgcl₂?

Explanation:

Step1: Calculate \(\%(m/m)\)

The formula for \(\%(m/m)\) is \(\frac{\text{mass of solute}}{\text{mass of solution}}\times100\%\).
Given mass of solute (\(NaCl\)) \(m = 3.26\space g\) and mass of solution \(M=15.78\space g\).
\(\%(m/m)=\frac{3.26\space g}{15.78\space g}\times 100\%\)
\(=\frac{326}{15.78}\% \approx 20.7\%\)

Step2: Calculate \(\%(m/v)\)

The formula for \(\%(m/v)\) is \(\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times100\%\).
Given mass of solute (\(NaCl\)) \(m = 3.26\space g\) and volume of solution \(V = 15.0\space mL\).
\(\%(m/v)=\frac{3.26\space g}{15.0\space mL}\times100\%\)
\(=\frac{326}{15}\% \approx 21.7\%\)

Step3: Calculate molarity (\(M\))

First, find the molar mass of \(NaCl\). The molar mass of \(Na\) is \(22.99\space g/mol\) and of \(Cl\) is \(35.45\space g/mol\). So, molar mass of \(NaCl,M_{NaCl}=22.99 + 35.45=58.44\space g/mol\).
Moles of \(NaCl,n=\frac{\text{mass of }NaCl}{\text{molar mass of }NaCl}=\frac{3.26\space g}{58.44\space g/mol}\approx0.0558\space mol\).
Volume of solution \(V = 15.0\space mL=0.0150\space L\).
The formula for molarity \(M=\frac{n}{V}\).
\(M=\frac{0.0558\space mol}{0.0150\space L}=3.72\space M\)

Answer:

a. \(20.7\%\)
b. \(21.7\%\)
c. \(3.72\space M\)