try this! solve for the following sample problem. please be guided with gfsa format and box your final…

try this! solve for the following sample problem. please be guided with gfsa format and box your final answer. 1. what is the molarity of a solution containing 25g of cuso₄·5h₂o in 350 ml of water? 2. consider an example in which 18g of c₆h₁₂o₆ is dissolved in 1kg of water. what is the molality of the resulting solution? 3. an industrial exhaust contains 85 g of nitrogen dioxide (no₂) mixed with 1200 g of carbon monoxide (co). what is the mole fraction of no₂ in the gas mixture?

try this! solve for the following sample problem. please be guided with gfsa format and box your final answer. 1. what is the molarity of a solution containing 25g of cuso₄·5h₂o in 350 ml of water? 2. consider an example in which 18g of c₆h₁₂o₆ is dissolved in 1kg of water. what is the molality of the resulting solution? 3. an industrial exhaust contains 85 g of nitrogen dioxide (no₂) mixed with 1200 g of carbon monoxide (co). what is the mole fraction of no₂ in the gas mixture?

Answer

Answer:

First, find the molar mass of $CuSO_4\cdot5H_2O$. The molar mass of $Cu = 63.55\ g/mol$, $S=32.07\ g/mol$, $O = 16.00\ g/mol$, $H=1.01\ g/mol$. For $CuSO_4\cdot5H_2O$, $M=(63.55 + 32.07+4\times16.00 + 5\times(2\times1.01+16.00))\ g/mol=(63.55 + 32.07+64.00+5\times18.02)\ g/mol=(63.55 + 32.07+64.00 + 90.10)\ g/mol = 249.72\ g/mol$. The number of moles of $CuSO_4\cdot5H_2O$, $n=\frac{m}{M}=\frac{25\ g}{249.72\ g/mol}\approx0.100\ mol$. The volume of the solution $V = 350\ mL=0.350\ L$. The molarity $M=\frac{n}{V}=\frac{0.100\ mol}{0.350\ L}\approx0.286\ M$.

The molar mass of $C_6H_{12}O_6$ is $M=(6\times12.01 + 12\times1.01+6\times16.00)\ g/mol=(72.06+12.12 + 96.00)\ g/mol = 180.18\ g/mol$. The number of moles of $C_6H_{12}O_6$, $n=\frac{m}{M}=\frac{18\ g}{180.18\ g/mol}\approx0.100\ mol$. The mass of the solvent is $m_{solvent}=1\ kg$. The molality $m=\frac{n}{m_{solvent}}=\frac{0.100\ mol}{1\ kg}=0.100\ m$.

The molar mass of $NO_2$ is $M_{NO_2}=(14.01+2\times16.00)\ g/mol = 46.01\ g/mol$. The number of moles of $NO_2$, $n_{NO_2}=\frac{m_{NO_2}}{M_{NO_2}}=\frac{85\ g}{46.01\ g/mol}\approx1.85\ mol$. The molar mass of $CO$ is $M_{CO}=(12.01 + 16.00)\ g/mol=28.01\ g/mol$. The number of moles of $CO$, $n_{CO}=\frac{m_{CO}}{M_{CO}}=\frac{1200\ g}{28.01\ g/mol}\approx42.84\ mol$. The mole - fraction of $NO_2$, $x_{NO_2}=\frac{n_{NO_2}}{n_{NO_2}+n_{CO}}=\frac{1.85\ mol}{1.85\ mol + 42.84\ mol}=\frac{1.85\ mol}{44.69\ mol}\approx0.041$.

Explanation:

Step1: Calculate molar mass

For each compound, sum atomic masses.

Step2: Calculate number of moles

Use $n=\frac{m}{M}$.

Step3: Calculate molarity/molality/mole - fraction

For molarity $M=\frac{n}{V}$, for molality $m=\frac{n}{m_{solvent}}$, for mole - fraction $x=\frac{n_i}{\sum n_i}$.