Solutions and Concentration - SS1 Chemistry Past Questions and Answers - page 4

Step 1: Convert the given mass of glucose to moles.

Molar mass of glucose (C6H12O6)

= (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol)

= 180.18 g/mol

Number of moles of glucose = Mass of glucose / Molar mass of glucose

= 120 g / 180.18 g/mol

≈ 0.666 mol

 

Step 2: Convert the given mass of water to kilograms.

Mass of water = 500 g = 500/1000 kg = 0.5 kg

 

Step 3: Calculate molality using the formula:

Molality (m) = Number of moles of solute / Mass of solvent (in kg)

= 0.666 mol / 0.5 kg

= 1.332 m

The molality of the solution is 1.332 m.

Step 1: Calculate the total mass of the solution.

Total mass of the solution = Mass of ethanol + Mass of water

= 30 g + 70 g = 100 g

Step 2: Calculate the mass percentage using the formula:

Mass percentage = (Mass of solute / Total mass of solution) x 100

= (30 g / 100 g) x 100

= 30%

The mass percentage of ethanol in the solution is 30%.

0.5 g

In a w/v (weight/volume) solution, the weight of the solute (glucose) is expressed in grams, and the volume of the solution is expressed in millilitres (mL).

 

Given:

Volume of the solution (V2) = 500 mL

Concentration of the solution (C2) = 10% (w/v)

Weight of the solute (glucose) = ?

 

To find the weight of glucose, we need to convert the percentage to a decimal. 10% = 10/100 = 0.1.

 

Weight of glucose = (Concentration × Volume) / 100

Weight of glucose = (0.1 × 500 mL) / 100

Weight of glucose = 0.5 g

150 mL

Using the dilution formula, we can calculate the volume of water (V2) to be added to the stock solution.

 

Given:

C1 (stock solution concentration) = 1 M

V1 (stock solution volume) = ?

C2 (diluted solution concentration) = 0.4 M

V2 (diluted solution volume) = 250 mL

 

Rearranging the dilution formula, we get:

V1 = (C2 × V2) / C1

Plugging in the values:

V1 = (0.4 M × 250 mL) / 1 M

V1 = 100 mL

 

To find the volume of water, subtract the stock solution volume from the desired diluted solution volume:

Volume of water = V2 - V1

Volume of water = 250 mL - 100 mL

 

Volume of water = 150 mL

To calculate the volume of the stock solution needed, we can use the dilution equation:

 

C1V1 = C2V2

 

Where:

C1 = initial concentration (10 M)

V1 = initial volume (unknown)

C2 = final concentration (2 M)

V2 = final volume (500 mL)

 

Rearranging the equation, we have:

 

V1 = (C2V2) / C1

 

V1 = (2 M x 500 mL) / 10 M

V1 = 100 mL

 

Therefore, you would measure 100 mL of the 10 M HCl stock solution and then add enough solvent (typically water) to make the total volume 500 mL.

Using the same dilution equation:

 

C1V1 = C2V2

 

Where:

C1 = initial concentration (2 M)

V1 = initial volume (unknown)

C2 = final concentration (0.5 M)

V2 = final volume (250 mL)

 

Rearranging the equation, we have:

 

V1 = (C2V2) / C1

 

V1 = (0.5 M x 250 mL) / 2 M

V1 = 62.5 mL

 

To prepare the 0.5 M NaOH solution, you would measure 62.5 mL of the 2 M NaOH stock solution and then add enough solvent to make the total volume 250 mL.

Question:

You have a 1 L solution of glucose with a concentration of 0.4 M. How would you prepare 500 mL of a 0.1 M glucose solution through dilution?

Answer:

Using the dilution equation:

 

C1V1 = C2V2

 

Where:

C1 = initial concentration (0.4 M)

V1 = initial volume (1 L or 1000 mL)

C2 = final concentration (0.1 M)

V2 = final volume (500 mL)

 

Rearranging the equation, we have:

 

V1 = (C2V2) / C1

 

V1 = (0.1 M x 500 mL) / 0.4 M

V1 = 125 mL

 

To prepare the 0.1 M glucose solution, you would measure 125 mL of the 0.4 M glucose solution and then add enough solvent to make the total volume 500 mL.