How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel
1 answer:
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Answer:
0.39 % m/m; 0.42 % m/v; 0.18 % v/v; 4200 ppm; 0.044 mol·L⁻¹; 0.041 mol/kg;
0.089 equiv/L; 0.000 74; 0.999 26
Explanation
Data:
Mass of MgCl₂ = 3.8 g
Volume of solution = 900 mL
Density of solution = 1.09 g/mL
Density of MgCl₂ = 2.32 g/cm³
Calculations
1. Percent m/m
2. Percent m/v
3. Percent v/v
4. Parts per million
5. Molar concentration
6. Molal concentration
Mass of water = Mass of solution - mass of solute = 981 g - 3.8 g = 977 g = 0.977 kg
7. Normality
The normality is the number of equivalents per litre of solution.
The normality of an ion equals the molar concentration times the charge on the ion.
Thus, the normality of MgCl₂ is twice the molar concentration.
Normality = 2 × 0.044 mol·L⁻¹ = 0.089 equiv·L⁻¹
8. Mole fraction of solute
Moles of MgCl₂ = 0.040 mol
Total moles = n₂ + n₁ = 0.040 mol + 54.23 mol = 54.27 mol
9. Mole fraction of solvent
χ₁ = 1 - χ₂ = 1 - 0.000 74 = 0.999 26