Question

Calculate the molality of each of the following solutions: (a) $583 \mathrm{g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $1.50 \mathrm{kg}$ of water $-$ the acid solution used in an automobile battery (b) $0.86 \mathrm{g}$ of $\mathrm{NaCl}$ in $1.00 \times 10^{2} \mathrm{g}$ of water $-$ a solution of sodium chloride for intravenous injection (c) 46.85 g of codeine, $\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}$, in 125.5 $\mathrm{g}$ of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (d) 25 g of I $_{2}$ in 125 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

Name: Problem 38, Chapter 11: Chemistry
Uploaded: 2019-09-02T23:11:15-08:00
Duration: 4 min 18 s
Channel: Rikhil Makwana
Description: Calculate the molality of each of the following solutions: (a) 583 g of H2 SO4 in 1.50 kg of water - the acid solution used in an automobile battery (b) 0.86 g of NaCl in 1.00 ×10^2 g of water - a solution of sodium chloride for intravenous injection (c) 46.85 g of codeine, C18 H21 NO3, in 125.5 g of ethanol, C2 H5 OH (d) 25 g of I 2 in 125 g of ethanol, C2 H5 OH

   Calculate the molality of each of the following solutions:
(a) $583 \mathrm{g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $1.50 \mathrm{kg}$ of water $-$ the acid solution used in an automobile battery
(b) $0.86 \mathrm{g}$ of $\mathrm{NaCl}$ in $1.00 \times 10^{2} \mathrm{g}$ of water $-$ a solution of sodium chloride for intravenous injection
(c) 46.85 g of codeine, $\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}$, in 125.5 $\mathrm{g}$ of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
(d) 25 g of I $_{2}$ in 125 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

Chemistry

Paul Flowers, Klaus… 2nd Edition

Chapter 11, Problem 38

Instant Answer

Solved by Expert Rikhil Makwana

09/02/2019

Step 1

To do this, we'll divide the mass of the solute by its molar mass. (a) For H2SO4, the molar mass is (2 x 1.01) + (32.07) + (4 x 16.00) = 98.09 g/mol. So, the number of moles of H2SO4 is: moles = 583 g / 98.09 g/mol = 5.94 mol (b) For NaCl, the molar mass is Show more…

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Calculate the molality of each of the following solutions: (a) $583 \mathrm{g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in $1.50 \mathrm{kg}$ of water $-$ the acid solution used in an automobile battery (b) $0.86 \mathrm{g}$ of $\mathrm{NaCl}$ in $1.00 \times 10^{2} \mathrm{g}$ of water $-$ a solution of sodium chloride for intravenous injection (c) 46.85 g of codeine, $\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}$, in 125.5 $\mathrm{g}$ of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ (d) 25 g of I $_{2}$ in 125 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

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Key Concepts

Molality

Molality is a concentration measure defined as the number of moles of solute per kilogram of solvent. It is particularly useful in calculations involving colligative properties because it remains unaffected by temperature changes, given that it is based on mass rather than volume.

Mole and Molar Mass

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. Converting a substance's mass to moles requires using its molar mass, which is the mass of one mole of that substance. This conversion is essential for determining the solute quantity in concentration calculations such as molality.

Unit Conversion

Accurate unit conversion is crucial in quantitative chemical calculations. Converting masses to the appropriate units (for example, grams to kilograms) ensures that formulas like the molality equation are applied correctly, thereby yielding accurate results.

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