Question

CHE 201 - SOLUTIONS, SOLUBILITIES AND CONCENTRATIONS Name Date Pre-Laboratory Questions: Submit these immediately upon arriving at the laboratory You must show ALL of your calculations to receive credit for a correct answer. 1. The solubility of sucrose (common table sugar) at 70. "C is 320 g/100g H,0. a. How many grams of sucrose can dissolve in 200. g of water at 70°C? b. Will 400 g of sucrose dissolve in a teapot that contains 200 g of water at 70°C? Explain. 2. A solution is prepared with 3.26 g KCl and water to make 25.0 mL of KCI. a. What is the % (m/v) of the KCl solution? b. What is the molarity (M) of the KCl solution?

Answer 1

Answer:

1) Solubility of sucrose is 320.0 g / 100.0 of water. This means maximum of 320.0 g sucrose can dissolve in 100.0 g of water solvent. We write a relationship is,

320.0 g of sucrose $\equiv$ 100.0 g of water. ............. (1)

a) Let "A" g sucrose dissolve in 200.0 g water. Hence we write,

If,   320.0 g of sucrose $\equiv$ 100.0 g of water.

Then, "A" g of sucrose $\equiv$ 200.0 g of water.

On cross multiplication,

A * 100 g = 320.0 g * 200.0 g

A = 320.0 g * 200.0 g / 100.0 g

A = 640.0 g

640.0 g of sucrose can be dissolved in 200.0 g water.

b) From above calculation for part (a) it's clear that maximum of 640.0 g of sucrose can be dissolved in 200.0 g water.

Hence its obvious that 400.0 g sucrose can easily dissolved in 200.0 g water.

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2)

a) Mass of KCl dissolved = 3.26 g. Volume of solution = 25.0 mL.

KCl (m/V%) = (Mass of KCl dissolved / Volume of Solution)*100 = (3.26 / 25.0 mL)*100 = 13.04 (m/V%).

b) To calculate molarity we need to calculate # of moles of KCl.

Mass of KCl = 3.26 g, Molar mass of KCl = 74.55 g.mol^-1.

# of moles of KCl = Mass of KCl / Molar mass of KCl

= 3.26 g / 74.55 g.mol^-1.

= 0.044 mol.

Volume of KCl solution = 25.0 mL = 0.025 L

Molarity of KCl = # of moles of KCl / Volume of solution in L = 0.044 mol / 0.025 L = 1.75 mol/L = 1.75 M.

Molarity of KCl = 1.75 M

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