Question

missed this? watch kcv: solution concentration, we: using molarity in calculations; read section 5.2. you can click on the review link to access the section in your etext. a laboratory procedure calls for making 500.0 ml of a 1.40 m k₂so₄ solution. part a what mass of k₂so₄ (in g) is needed? express the mass in grams to three significant figures. view available hint(s)

Explanation:

Step1: Calculate the number of moles of \(K_2SO_4\)

The formula for molarity \(M=\frac{n}{V}\), where \(M\) is molarity, \(n\) is the number of moles, and \(V\) is the volume in liters. First, convert the volume \(V = 500.0\ mL=0.5000\ L\) and \(M = 1.40\ M\). Rearranging the formula for \(n\), we get \(n = M\times V\). So \(n=1.40\ mol/L\times0.5000\ L = 0.700\ mol\).

Step2: Calculate the molar - mass of \(K_2SO_4\)

The molar - mass of \(K\) is approximately \(39.10\ g/mol\), \(S\) is approximately \(32.07\ g/mol\), and \(O\) is approximately \(16.00\ g/mol\). For \(K_2SO_4\), \(M_{K_2SO_4}=2\times39.10\ g/mol + 32.07\ g/mol+4\times16.00\ g/mol=174.27\ g/mol\).

Step3: Calculate the mass of \(K_2SO_4\)

The formula for mass \(m=n\times M\). Substitute \(n = 0.700\ mol\) and \(M = 174.27\ g/mol\) into the formula. So \(m=0.700\ mol\times174.27\ g/mol = 122\ g\) (rounded to three significant figures).

Answer:

122 g