Math  /  Numbers & Operations

QuestionCalculate the molar concentration of each solution. a. 14 g of copper(II) sulfate, CuSO4( s)\mathrm{CuSO}_{4}(\mathrm{~s}), dissolved in 70 mL of solution b. 5.07 g of sucrose, C12H22O11( s)\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{~s}), dissolved in 23.6 mL of solution c. 1.1 g of calcium nitrate, Ca(NO3)2( s)\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}), dissolved in 70 mL of solution

Studdy Solution

STEP 1

1. Molar concentration (molarity) is defined as the number of moles of solute per liter of solution.
2. We need the molar mass of each compound to convert grams to moles.
3. The volume of the solution should be converted from milliliters to liters.

STEP 2

1. Calculate the molar mass of each compound.
2. Convert the mass of each solute to moles.
3. Convert the volume of the solution to liters.
4. Calculate the molarity of each solution.

STEP 3

Calculate the molar mass of copper(II) sulfate, CuSO4\mathrm{CuSO}_{4}.
- Copper (Cu): 63.55g/mol63.55 \, \text{g/mol} - Sulfur (S): 32.07g/mol32.07 \, \text{g/mol} - Oxygen (O): 16.00g/mol×4=64.00g/mol16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}
Molar mass of CuSO4=63.55+32.07+64.00=159.62g/mol\text{Molar mass of } \mathrm{CuSO}_{4} = 63.55 + 32.07 + 64.00 = 159.62 \, \text{g/mol}

STEP 4

Convert 14g14 \, \text{g} of CuSO4\mathrm{CuSO}_{4} to moles.
Moles of CuSO4=14g159.62g/mol=0.0877mol\text{Moles of } \mathrm{CuSO}_{4} = \frac{14 \, \text{g}}{159.62 \, \text{g/mol}} = 0.0877 \, \text{mol}

STEP 5

Convert 70mL70 \, \text{mL} to liters.
70mL=0.070L70 \, \text{mL} = 0.070 \, \text{L}

STEP 6

Calculate the molarity of the CuSO4\mathrm{CuSO}_{4} solution.
Molarity=0.0877mol0.070L=1.253M\text{Molarity} = \frac{0.0877 \, \text{mol}}{0.070 \, \text{L}} = 1.253 \, \text{M}

STEP 7

Calculate the molar mass of sucrose, C12H22O11\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}.
- Carbon (C): 12.01g/mol×12=144.12g/mol12.01 \, \text{g/mol} \times 12 = 144.12 \, \text{g/mol} - Hydrogen (H): 1.01g/mol×22=22.22g/mol1.01 \, \text{g/mol} \times 22 = 22.22 \, \text{g/mol} - Oxygen (O): 16.00g/mol×11=176.00g/mol16.00 \, \text{g/mol} \times 11 = 176.00 \, \text{g/mol}
Molar mass of C12H22O11=144.12+22.22+176.00=342.34g/mol\text{Molar mass of } \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} = 144.12 + 22.22 + 176.00 = 342.34 \, \text{g/mol}

STEP 8

Convert 5.07g5.07 \, \text{g} of sucrose to moles.
Moles of sucrose=5.07g342.34g/mol=0.0148mol\text{Moles of sucrose} = \frac{5.07 \, \text{g}}{342.34 \, \text{g/mol}} = 0.0148 \, \text{mol}

STEP 9

Convert 23.6mL23.6 \, \text{mL} to liters.
23.6mL=0.0236L23.6 \, \text{mL} = 0.0236 \, \text{L}

STEP 10

Calculate the molarity of the sucrose solution.
Molarity=0.0148mol0.0236L=0.627M\text{Molarity} = \frac{0.0148 \, \text{mol}}{0.0236 \, \text{L}} = 0.627 \, \text{M}

STEP 11

Calculate the molar mass of calcium nitrate, Ca(NO3)2\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}.
- Calcium (Ca): 40.08g/mol40.08 \, \text{g/mol} - Nitrogen (N): 14.01g/mol×2=28.02g/mol14.01 \, \text{g/mol} \times 2 = 28.02 \, \text{g/mol} - Oxygen (O): 16.00g/mol×6=96.00g/mol16.00 \, \text{g/mol} \times 6 = 96.00 \, \text{g/mol}
Molar mass of Ca(NO3)2=40.08+28.02+96.00=164.10g/mol\text{Molar mass of } \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} = 40.08 + 28.02 + 96.00 = 164.10 \, \text{g/mol}

STEP 12

Convert 1.1g1.1 \, \text{g} of calcium nitrate to moles.
Moles of calcium nitrate=1.1g164.10g/mol=0.0067mol\text{Moles of calcium nitrate} = \frac{1.1 \, \text{g}}{164.10 \, \text{g/mol}} = 0.0067 \, \text{mol}

STEP 13

Convert 70mL70 \, \text{mL} to liters.
70mL=0.070L70 \, \text{mL} = 0.070 \, \text{L}

STEP 14

Calculate the molarity of the calcium nitrate solution.
Molarity=0.0067mol0.070L=0.096M\text{Molarity} = \frac{0.0067 \, \text{mol}}{0.070 \, \text{L}} = 0.096 \, \text{M}
The molar concentrations are: a. 1.253MCuSO41.253 \, \text{M} \, \mathrm{CuSO}_{4} b. 0.627MC12H22O110.627 \, \text{M} \, \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} c. 0.096MCa(NO3)20.096 \, \text{M} \, \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}

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