Question

5 moles o glucose is dissolved in 1.1 kg of water if the density of solution is 1.25 gm/ml then determine %w/v if the solution.

Answer 1

Explanation:

Here's how you can go about solving this one.

Explanation:

The problem gives you all the information you need in order to solve for the molality and mole fraction of the solution. In order to determine its molarity, you're going to need the solution's volume.

To get the volume, you have to know what the density of the solution is. Determine the percent concentration by mass of the solution first

%w/w

=

m

solute

m

solution

⋅

100

In your case, the mass of the solution will be

m

solution

=

m

glucose

+

m

water

m

solution

=

20

+

150

=

170 g

This means that you get

%w/w

=

20

g

170

g

⋅

100

=

11.8%

ρ

=

1.045 g/mL

Use glucose's molar mass to determine how many moles you have

20

g

⋅

1 mole glucose

180.16

g

=

0.111 moles glucose

The solution's volume will be

170

g

⋅

1 mL

1.045

g

=

162.7 mL

C

=

n

V

=

0.111 moles

162.7

⋅

10

−

3

L

=

0.68 M

b

=

n

m

water

=

0.111 moles

150

⋅

10

−

3

kg

=

0.74 molal

150

g

⋅

1 mole water

18.02

g

=

8.24 moles water

The total number of moles the solution contains is

n

total

=

n

glucose

+

n

water

n

total

=

0.111

+

8.24

=

8.351 moles

χ

sucrose

=

n

sucrose

n

total

=

0.111

moles

8.351

moles

=

0.013

SIDE NOTE I've left the values rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of glucose.

Answer 2

The %w/v of glucose in the solution is 56.25%.

Given,

Moles of glucose=5

Mass of water(solvent)=1.1 Kg

Density of the solution=1.25 g/mL.

To find,

the %w/v if the solution.

Solution:

• Weight by volume percentage of a substance in a solution is defined as the mass of the substance in 100 mL of the solution.
• Weight by volume percentage of a substance is temperature dependent.

%w/v=(Mass of the substance/Volume of the substance(mL)) x 100

• The molar mass of glucose is 180 g.

Moles=(Given mass/Molar mass)

$5=\frac{Mass}{180} \\Mass=180X5\\Mass=900 g.$

900 g of glucose is present in the solution.

Mass of water=1.1 Kg

$1.1Kg=1.1X1000 g\\1.1 Kg=1100 g.$

Mass of the solution=(900+1100)Kg

Mass of the solution is 2000 g.

$Density=\frac{Mass}{Volume} \\1.25=\frac{2000}{Volume} \\Volume=\frac{2000}{1.25} \\Volume=1600 mL.$

Volume of the solution is 1600 mL.

%w/v=(Mass of the substance/Volume of the substance(mL))x100

$w/v=\frac{900}{1600} X100\\w/v=\frac{900}{16} \\w/v=56.25$

The %w/v of the solution is 56.25%.

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