Nandita needs to prepare a Ringer\'s solution before she can perform an experime

Question

Nandita needs to prepare a Ringer's solution before she can perform an experiment. Nandita wants to prepare a concentrated stock solution of Ringer's which she can dilute to a working concentration before each experiment begins. Nandita wants the stock solution to be ten times more concentrated than the working solution she will use for experiments. The Ringer's solution she uses for experiments is 147 mM NaCI, 0.004 M KC1, and 0.003 M CaCla. Describe using a combination of numbers and words how Nandita would prepare 500 mL of the concentrated stock Ringer's solution. Make sure to include the mass amount of each compound Nandita must measure out. 3. 4. What wavelength of light will be emitted when an electron in hydrogen falls from an n- 2 to a n= 1?

Explanation / Answer

Q3)

Concentration of stock solution = 10 * concentration of experiment solution

Concentration of NaCl in stock solution = 1470mM NaCl solution = 1.47M solution

Concentration of KCl in stock solution = 0.04M KCl solution

Concentration of CaCl2 in stock solution = 0.03M CaCl2 solution

Since we need to create total volume of 500 mL solution

Moles of NaCl required = Volume of solution in L * Molarity = 500/1000 * 1.470 = 0.735 moles

Moles of KCl required = Volume of solution in L * Molarity = 500/1000 * 0.04 = 0.02 moles

Moles of CaCl2 required = Volume of solution in L * Molarity = 500/1000 * 0.03 = 0.015 moles

Mass of NaCl required = number of moles * molar mass = 0.735 mol * (23+35.5) gm/mol = 42.9975 grams of NaCl

Mass of KCl required = number of moles * molar mass = 0.02 mol * (39+35.5) gm/mol = 1.49 grams of KCl

Mass of CaCl2 required = number of moles * molar mass = 0.015 mol * (40+2*35.5) gm/mol = 1.665 grams of CaCl2

4)

Energy of electron in hydrogen atom = -13.6/n^2 eV

For n=2 to n=1 transition

Energy = 13.6/2^2 - 13.6/1^2 = 10.2 eV = 10.2 * 1.6 * 10^(-19) = 1.632 * 10^(-18)

Energy = hc/lambda

lambda = (6.626 * 10^(-34) * 3 * 10^8)/(1.632 * 10^(-18)) = 122 nm

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