Question

Calculate the mole fraction of NaOH in 10% w/w aqueous solution. [At. MassH=1, O=16, Na=23u]​

Answer 1

Here if an 100 g of an diluted NaOH solution is marked as 10%(w/w) of NaOH then W

2

​

(mass of solute)=10g and mass of water(W

1

​

)=90g.

density=

V

solution

​

mass

​

V

solution

​

=

1.070

100

​

=93.46cm

−3

=0.09346L

number of moles of NaOH(n

solute

​

)=

49g(mol)

−1

10g

​

=0.2mol

Molarity=

V

solution

​

n

solute

​

​

Molarity=

0.09346L

0.2mol

​

=2.14M

molality=

W

1

​

(inkg)

n

solute

​

​

molality=

0.09kg

0.2mol

​

=2.22m

Let mole fraction of water and NaOH be x

1

​

 and x

2

​

and n

1

​

 and n

2

​

 be the moles of water and NaOH

n

1

​

=0.2mol and n

2

​

=

18g(mol)

−1

90g

​

=5mol

x

1

​

=

n

1

​

+n

2

​

n

1

​

​

=0.96

x

2

​

=

n

1

+n

2

n

2

​

=0.04

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Answer 2

Explanation:

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