Question

An aques solution of NaCl is prepared by
dissolving 25 kg of NaCl in 100kg of water.
Determine weight % & mole % of both NaCl & water??​

Answer 1

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As we know that,

Density=

Volume

Mass

Given:-

Density of solution =1.0816gm/mL

Volume of solution =100mL

∴1.0816=

100

Mass of solution

⇒Mass of solution=108.16g

Mass of solute (Na

2

CO

3

)=8.653g

Therefore,

Mass of solvent =108.16−8.653=99.507g

Molecular mass of Na

2

CO

3

=83g

As we know that,

No. of moles =

Mol. mass

Given mass

∴ No. of moles of Na

2

CO

3

=

83

8.653

=0.104 mol

Now,

As we know that,

Molarity =

Volume of solution

(in L)

No. of moles of solute

Therefore,

Molarity =

100

0.104

×1000=1.04M

Again, as we know that,

Molality =

Mass of solution

(in Kg)

No. of moles of solute

Therefore,

Molality =

99.507

0.104

×1000=1.045m

Hence the molarity and molality of solution are 1.04M and 1.045m respectively.

Answer 2

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As we know that,

Density=

Volume

Mass

Given:-

Density of solution =1.0816gm/mL

Volume of solution =100mL

∴1.0816=

100

Mass of solution

⇒Mass of solution=108.16g

Mass of solute (Na

2

CO

3

)=8.653g

Therefore,

Mass of solvent =108.16−8.653=99.507g

Molecular mass of Na

2

CO

3

=83g

As we know that,

No. of moles =

Mol. mass

Given mass

∴ No. of moles of Na

2

CO

3

=

83

8.653

=0.104 mol

Now,

As we know that,

Molarity =

Volume of solution

(in L)

No. of moles of solute

Therefore,

Molarity =

100

0.104

×1000=1.04M

Again, as we know that,

Molality =

Mass of solution

(in Kg)

No. of moles of solute

Therefore,

Molality =

99.507

0.104

×1000=1.045m

Hence the molarity and molality of solution are 1.04M and 1.045m respectively.