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Molality Quiz 2

chemistry

 Calculate the molality of a 5.5 percent nitric acid solution with a density of $1.04 g/mL. $

 The molality can also be determined using a straightforward method of determining the mass of the solvent without the use of density data.

 $5.5\%$ of nitric acid means 5.5g of nitric acid is present in 100g of solution. Thus, mass of water is given as difference in mass of solution and mass of the solute and that is, 100-5.5=94.5g. Molality of this solution will be = $\dfrac{m}{{{M}_{HN{{O}_{3}}}}\times {{M}_{solvent}}}=\dfrac{5.5}{63\times 94.5}\times 1000=0.923m$ 

 Priya made a sugar solution by combining $10g$ of sugar cube with $500ml$ of water heated to $80^{\circ} C$ in a teacup. What will be the molality of this solution prepared by Priya? (density of water at ${{80}^{0}}C=0.975g/ml$

 The sugar to be considered here is the sucrose which is the common sugar we use for sweetening.

 The sugar has the molecular formula ${{C}_{12}}{{H}_{22}}{{O}_{11}}$ with the molar mass of 342g/mol. We know that density is the ratio of mass and volume and hence we can rearrange this to finding mass as, $m=d\times V$ Hence, mass of sugar will be,  $m=0.975\times 500=487.5g$ Now, number of moles of solute is the ratio of mass of solute to that of molar mass and that is, $\dfrac{10}{342}=0.0292mol.$ Now, molality is $\dfrac{0.0292}{487.5}\times 1000=0.059m $ 

 What is the molality of a solution containing $5.65g$ of glucose dissolved in $25.0g$ of water, resulting in the formation of blood sugar?

 Any given solution's molality is also known as a molal solution, and it describes the amount of moles of that solute per kilogram of solvent. The molality of glucose in blood is simply the amount of glucose present, which can be calculated using a simple molality formula.

 Molar mass of glucose is 180g/mol. Number of moles of glucose will be, $\dfrac{mass}{Mol.mass}=\dfrac{5.65g}{180g/mol}=0.03139mol\\$ Now, 25g of water contains 0.03139 mol of glucose. Thus, 1000g of water contains $m=\dfrac{0.013139}{25}\times 1000=1.25m$ 

 The electrolyte solution in a lead storage battery is $3.95M$ sulphuric acid solution with a density of $1.53g/mL$. Find the molality of this acid solution.<br>A) $3.533m$<br>B) $3.562m$<br>C) $3.566m$<br>D) $3.466m$

 Molality is a measure of the number of moles of solute present in 1 kg of solvent. 

 Now, the density of solution is 1.53g/mL. which is nothing but 1530g/L. $\\$ 3.95M sulphuric acid contains 3.95 moles of ${{H}_{2}}S{{O}_{4}}$ per litre of solution that is $3.95\times 98=387.1g$ Thus, amount of solvent that is water present in 1 litre of solution is =1530-387.1=1142.9g. Thus, molality of the solution will be, m=$\dfrac{3.95}{1142.9}\times 1000=3.456m$ 

 How can you determine the molality of ethanol in a solution made by dissolving $1.0g$ of ethanol in $100.0g$ water and obtaining a final volume of $101.0 mL$?

 Molality can be conveniently measured using the mass of the solute and the known molar mass of the solute by dividing their ratio by the volume of solvent used.

 Number of moles of ethanol is found by taking the ratio of mass of ethanol and molecular mass of ethanol. Molecular mass of ethanol is 46g/mol. Thus, the number of moles of ethanol is found to be, $\dfrac{1g}{46g/mol}=0.0217mol$ Mass of water is given as 100g. Now, molality will be =$\dfrac{0.0217}{100}\times 1000=0.217m$ 