please help with part C)show work Section: Objective #11- Students can discuss the formation and properties solutions and cansolve problems involving solutions.1. a. A solution is created by mixing 35.3g of C,H12O6 with 100.0mL of water. (Molar Massof C6H1206 = 180.18 g/mol). What is the molarity of the solution created assuming thevolume of the water is the final volume of the solution as well?b. Another solution is created by adding 75.0mL of water to 25.0mL of 1.50M NaCl. What isthe molarity of this new solution?c. Which solution, the C,H12O6 (a) or the NaCl (b), will have the highest boiling point? Explainyour choice.

Given data: Mass of C6H12O6 = 35.3 g Volume of water = 100 mL Molar mass of C6H12O6 = 180.18g/mol

Answered: 1. a. A solution is created by mixing…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

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please help with part C)

show work

Section: Objective #11- Students can discuss the formation and properties solutions and can
solve problems involving solutions.
1. a. A solution is created by mixing 35.3g of C,H12O6 with 100.0mL of water. (Molar Mass
of C6H1206 = 180.18 g/mol). What is the molarity of the solution created assuming the
volume of the water is the final volume of the solution as well?
b. Another solution is created by adding 75.0mL of water to 25.0mL of 1.50M NaCl. What is
the molarity of this new solution?
c. Which solution, the C,H12O6 (a) or the NaCl (b), will have the highest boiling point? Explain
your choice.

Expert Solution

Step 1

Given data:

Mass of C₆H₁₂O₆ = 35.3 g

Volume of water = 100 mL

Molar mass of C₆H₁₂O₆ = 180.18g/mol

Step 2

(a)

The molarity of the solution is calculated as follows:

It is given that the volume of water is assumed to be the final volume of the solution.

Number of moles of C₆H₁₂O₆ is,

$\begin{array}{rcl} n_{C_{6} H_{12} O_{6}} & = & \frac{35.3 g}{180.18 \frac{g}{mol}} \\ = & 0.196 mol \end{array}$

Molarity is the ratio number of moles of the solute to the volume of solution in L.

It is represented by M.

$Molarity = \frac{Number of moles of solute}{Volume of solution in L}$ ............... (1)

Step 3

Plugin the values in equation (1)

$\begin{array}{rcl} Molarity & = & \frac{0.196 mol}{100 mL} \\ 1 L & = & 1000 mL \\ Molarity & = & \frac{0.196 mol}{100 mL \times \frac{1 L}{1000 mL}} \\ = & 1.96 \frac{mol}{L} \end{array}$

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