Question

rank the following solutions from increasing to decreasing molarity.
highest concentration
a solution containing 3 moles of bacl2 in 100 ml water
a solution containing 4 grams of licl in 2 liters of water
a solution containing 60 grams of nacl in 250 milliliters of water
lowest concentration

Explanation:

Step1: Calculate molarity of \(BaCl_2\) solution

Molarity formula \(M=\frac{n}{V}\) (where \(n\) is moles, \(V\) is volume in liters). Given \(n = 3\) moles, \(V=100\space mL=0.1\space L\).
\(M_{BaCl_2}=\frac{3}{0.1}=30\space M\)

Step2: Calculate molarity of \(LiCl\) solution

Molar mass of \(LiCl\): \(M_{LiCl}=6.94 + 35.45=42.39\space g/mol\). Moles of \(LiCl\), \(n=\frac{m}{M}=\frac{4}{42.39}\approx0.0944\space mol\). Volume \(V = 2\space L\).
\(M_{LiCl}=\frac{0.0944}{2}=0.0472\space M\)

Step3: Calculate molarity of \(NaCl\) solution

Molar mass of \(NaCl\): \(M_{NaCl}=22.99+35.45 = 58.44\space g/mol\). Moles of \(NaCl\), \(n=\frac{m}{M}=\frac{60}{58.44}\approx1.027\space mol\). Volume \(V=250\space mL = 0.25\space L\).
\(M_{NaCl}=\frac{1.027}{0.25}=4.108\space M\)

Answer:

Highest concentration: A solution containing 3 moles of \(BaCl_2\) in 100 mL water (\(30\space M\))
Middle: A solution containing 60 grams of \(NaCl\) in 250 milliliters of water (\(4.108\space M\))
Lowest concentration: A solution containing 4 grams of \(LiCl\) in 2 liters of water (\(0.0472\space M\))