Question

Part 3. Problems 1) The molarity of naphthaline in a solution of naphthalene (CH. MW-128.8), a non-ionizing and nonvolatile solute, and benzene (C.HS, MW = 78.11), a volatile solvent, is [CHx] = 0.125 M. The density of the solution is D= 0.8781 g/cm! Find the following: a) The osmotic pressure exerted by the solution, at T = 25.0°C. Give your final answer in units of torr. [8 points] b) Xn, the mole fraction of naphthalene in the solution. [15 points) 2) Thermochemical data are given below (at T = 25.0 °C) and may be useful in doing the following problem. Substance ΔΗ: (kJ/mol) S° (J/mol.K) AG°f (kJ/mol) - 466.9 Mg(aq) Mg(OH)2(8) OH(aq) - 924.7 - 230,0 - 138.1 63.1 - 10.8 -454.8 - 833.9 - 1572 Using only the data above, find AGºrn, AS, and K for the reaction Mg(OH)2(s) = Mg (aq) + 2 OH (aq) [18 points] 3) Hypobromous acid (HOBr) is a weak monoprotic acid, with K.-2.1 x 10" a) What is the value for pH for a 0.0424 M aqueous solution of hypobromous acid? [12 points) b) 0.0100 moles of NaOH, a strong soluble base, is added to 1.000 L of the above solution of hypobromous acid. What will be the pH for this new solution? [15 points)

Answer 1

Molarity: Molarity is a quantitative property of a chemical solution that informs about the content of solute in that solution. It is defined as the stoichiometry of solute against the total volume of that solution. Molarity is a measure of the concentration of solution. It is also helpful in identification of many other quantitative properties of the given and other associated solutions with which it is made to react.

Ans 1. a) Molarity of naphthalene in solution is 0.125 M

Given temperature is 25^oC or 298 K.

Osmotic pressure of a solution can be calculated by using the give formula.

$\pi =MRT$

substituting the values in the given formula,

$\pi =0.125\times 62.36\times 298$

$\pi =2322.91 Torr$

Therefore, the osmotic pressure of nathlene in the solution is 2322.9 Torr

b) Given density of total solution is 0.8781 g/L.

Mass of total solution = 1L* (0.8781 / 1 mL) * (1000 mL / 1 L)

Mass of total solution = 878.1 g

Number of moles of naphthalene can be calculated by the given method.

Number of moles naphthalene = 1L* ( 0.125 moles / 1L)

Number of moles naphthalene = 0.125 moles

Mass of naphthalene = number of moles * Molar mass

Mass of naphthalene = 0.125 * 128.17

Mass of naphthalene = 16.02 g

Mass of benzene = (878.1 - 16.02) g

Mass of benzene = 862.08 g

Number of moles of benzene = 862.08 / 78.11

Number of moles of benzene = 11.03

Mole fraction of naphthalene is solution is given by the given formula.

$Mole fraction of naphthalene (\chi _{N})= moles of naphthalene/(moles of naphthalene+moles of benzene)$

$Mole fraction of naphthalene (\chi _{N})= 0.125/(0.125+11.03)$

$Mole fraction of naphthalene (\chi _{N})= 0.011$

Mole fraction of naphthalene in the solution is 0.011