A2.750×10−2m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.3 ml . the density of water at 20.0∘c is 0.9982 g/ml.
calculate the mole fraction of salt in this solution.
calculate the concentration of the salt solution in percent by mass.
calculate the concentration of the salt solution in parts per million.
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Ответ:
Mole fraction of salt is 0.00049.
The concentration of the salt solution in percent by mass is 0.16%.
The concentration of the salt solution in parts per million is 1,610.18 .
Explanation:
Molarity of the NaCl solution =![2.750\times 10^{-2} M](/tpl/images/0174/2838/30556.png)
Moles of NaCl =![n_2](/tpl/images/0174/2838/7220e.png)
Volume of the solution = 1.000 L
Molarity=![\frac{\text{Moles of compound}}{\text{Volume of solution(L)}}](/tpl/images/0174/2838/10920.png)
Mass of
of NaCl :
Mass of water = m
Density of water = 0.9982 g/mL
Volume of water = 999.3 mL
Moles of water =![n_1=\frac{997.5 g}{18 g/mol}=55.416 mol](/tpl/images/0174/2838/b0632.png)
Mole fraction of salt =![\chi_1](/tpl/images/0174/2838/0ddc9.png)
Percentage by mass:
The concentration of the salt solution in percent by mass is 0.16%.
The concentration of the salt solution in parts per million.
Ответ:
"electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. In equation form, the electric potential difference is.."
check the link below. very helpful!
Explanation:
https://www.physicsclassroom.com/class/circuits/Lesson-1/Electric-Potential-Difference