Question

a 11.164 g mass of ethylene glycol (c₂h₆o₂, automotive antifreeze) is added to a 25 - ml volumetric flask. then, the flask is filled to the line (25.00 ml) with water to create an ethylene glycol solution. what is the molarity of the ethylene glycol solution?

Explanation:

Step1: Calculate the molar mass of ethylene glycol

The molar mass of \(C_2H_6O_2\) is \(M=(2\times12.01)+(6\times1.008)+(2\times16.00)=62.07\space g/mol\)

Step2: Calculate the number of moles of ethylene glycol

Using the formula \(n = \frac{m}{M}\), where \(m = 11.164\space g\) and \(M = 62.07\space g/mol\), we have \(n=\frac{11.164}{62.07}\approx0.18\space mol\)

Step3: Convert the volume of the solution to liters

The volume \(V = 25.00\space mL=25.00\times10^{- 3}\space L\)

Step4: Calculate the molarity

Using the formula \(Molarity(M)=\frac{n}{V}\), substituting \(n = 0.18\space mol\) and \(V=25.00\times 10^{-3}\space L\), we get \(M=\frac{0.18}{25.00\times10^{-3}} = 7.2\space mol/L\)

Answer:

\(7.2\space M\)