Question

what is the molarity of a solution of 49.1 g of calcium chloride (cacl2) in 230 ml of solution? report your answer to two places past the decimal point.

Explanation:

Step1: Calculate the molar mass of \(CaCl_2\)

The molar mass of \(Ca\) is \(40.08\space g/mol\), and the molar mass of \(Cl\) is \(35.45\space g/mol\). For \(CaCl_2\), \(M = 40.08+(2\times35.45)= 110.98\space g/mol\)

Step2: Calculate the number of moles of \(CaCl_2\)

Use the formula \(n=\frac{m}{M}\), where \(m = 49.1\space g\) and \(M = 110.98\space g/mol\). So \(n=\frac{49.1}{110.98}\approx0.442\space mol\)

Step3: Convert the volume of the solution to liters

Given \(V = 230\space mL\), since \(1\space L=1000\space mL\), then \(V = 230\div1000 = 0.23\space L\)

Step4: Calculate the molarity

Use the formula \(C=\frac{n}{V}\), where \(n = 0.442\space mol\) and \(V = 0.23\space L\). So \(C=\frac{0.442}{0.23}\approx1.92\space mol/L\)

Answer:

\(1.92\)