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Chapter 13: Problem 27

How many grams of iodine, \(I_{2}\), must be dissolved in \(725 \mathrm{mL}\) of carbon disulfide, \(\mathrm{CS}_{2}(d=1.261 \mathrm{g} / \mathrm{mL}),\) to produce a \(0.236 \mathrm{m}\) solution?

Short Answer

Expert verified
Approximately 54.763 grams of iodine must be dissolved in the carbon disulfide to produce a 0.236 m solution.

Step by step solution

01

Calculate the mass of the solvent

First, find the mass of the carbon disulfide (\(CS_2\)). The mass (\(m\)) can be calculated from the density (\(d\)) and volume (\(V\)) using the following equation: \(m = d \cdot V\). Substitute \(d = 1.261 g/mL\) and \(V = 725 mL\): \(m = (1.261 g/mL) * (725 mL) = 914.225 g\). Convert the mass from grams to kilograms by dividing by 1000: \(914.225 g = 0.914225 kg\).
02

Use the molality equation

The equation for molality (\(m\)) is defined as the number of moles of solute (\(n_s\)) divided by the mass of solvent (\(m_s\)) in kilograms: \(m = n_s/m_s\). Rearrange it to solve for \(n_s\): \(n_s = m \cdot m_s\). The given molality is \(0.236 m\), and from Step 1, we know that \(m_s = 0.914225 kg\). So, \(n_s = (0.236 m) * (0.914225 kg) = 0.2157771 moles\).
03

Convert moles to grams

Finally, convert the moles of iodine to grams using its molar mass. The molar mass of iodine (\(I_2\)) is \(253.80894 g/mole\). Thus, the mass of iodine can be calculated by multiplying the number of moles by the molar mass: \(Mass = n_s \cdot Molar\,Mass = (0.2157771 moles) * (253.80894 g/mole) = 54.763 g\). This means that one must dissolve approximately 54.763 grams of iodine to produce the desired solution.

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Most popular questions from this chapter

Nitrobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2},\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) are completely miscible in each other. Other properties of the two liquids are nitrobenzene: \(\mathrm{fp}=5.7^{\circ} \mathrm{C}, K_{\mathrm{f}}=\) \(8.1^{\circ} \mathrm{C} m^{-1} ;\) benzene: \(\mathrm{fp}=5.5^{\circ} \mathrm{C}, K_{\mathrm{f}}=5.12^{\circ} \mathrm{C} m^{-1} . \mathrm{It}\) is possible to prepare two different solutions with these two liquids having a freezing point of \(0.0^{\circ} \mathrm{C}\) What are the compositions of these two solutions, expressed as mass percent nitrobenzene?

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