Question

1. The concentrated sulfuric acid we use in the laboratory is 98.0 percent H2SO4 by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83g/mL

2. In a biochemical assay, a chemist needs to add 2.93 g of glucose to a reaction mixture. Calculate the volume in mL of a 1.35M glucose solution she should use for the addition.

Answer 1

1. The mass of 1 L of solution is calculated from its density.

$(1 \;L \;solution \times \frac{10^3 \;ml}{1\;L}) \times \frac{1.83 \;g \; solution}{mL \;solution} = 1830 \; g \; solution$

Then the mass of solute (H₂SO₄) present in 1 L of solution is determined from percent by mass:

$(\frac{1830 \;g \;solution}{L \; solution}) (\frac{98.0 \; g H_2SO_4}{100 \;g \; solution}) = \frac{1793 \;g \;H_2SO_4}{L\; solution}$

The molar mass of H₂SO₄ is 98.08 g/mol.

Now we divide mass of solute (H₂SO₄) present in 1 L of solution by molar mass and obtain molarity.

$\frac{1793 \;g \; H_2SO_4}{L \;solution} \times \frac{1 \;mol \;H_2SO_4}{98.08 \;g \;H_2SO_4} = \frac{18.3\; mol \;H_2SO_4}{L \; solution} = 18.3 \;M$

So, molarity of the solution is 18.3 M.

The mass of 1 L of solution is 1830 g and mass of solute (H₂SO₄) present in 1 L of solution is 1793 g. Then mass of solvent (water) in the solution is the difference between the mass of solution and mass of solute (H₂SO₄).

$(1830 \;g\; solution -1793 \;g \;H_2SO_4) = 37.0 \;g\;H_2O \times \frac{1 \;kg}{10^3 \;g} \\ = 0.0370 \;kg \;H_2O$

Molality of a solution is defined as the number of moles of solute dissolved in 1 kg of solvent.

$Molality \;(m) = \frac{moles \;of \;H_2SO_4}{mass \; of \;water \;(in \;kg)} \\ = \frac{18.3 \;mol \;H_2SO_4}{0.0370 \;kg\;H_2O} \\ = 495 \; m$

So, the molality of the solution is 495 m.

2. M(glucose) = 180 g/mol

n(glucose) $= \frac{2.93}{180} = 0.01627 \;mol$

Proportion:

1.35 mol – 1000 mL

0.01627 mol – x

$x=\frac{0.01627 \times 1000}{1.35} = 12.0 \;mL$

Answer: 12.0 mL