Question

he mole fraction of NaOH in 10% w/w aqueous solution of NaOH is 1/x
. The value of x is

Answer 1

Answer:

0.47. Hope my answer help you

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NaOH solution is prepared in water.

Molar mass of NaOH=40 g

Mass of NaOH=10 g

Molar mass of water = 18 g

Given mass of water = 90 g

Number of moles of NaOH=

40

10

=0.25 moles

Number of moles of H

2

O=

18

90

=5 moles

Mole fraction of NaOH=

0.25+5

0.25

=0.047

Answer 2

Answer:

The value of x is equal to 21.28.

Explanation:

Given: the solution of sodium hydroxide is prepared in water.

10% w/w aqueous solution of NaOH means 10g of NaOH in 100g of water.

Therefore, mass of NaOH $= 10g$

Mass of water (solvent) $=100-10=90g$

Molar mass of NaOH $= 23+16+1=40g$

And, molar mass of water $=18g$

Number of moles of NaOH = mass/molar mass $n_{1}=\frac{10}{40}=0.25moles$

Number of moles of water $n_{2}=\frac{90}{18}=5moles$

Mole fraction of NaOH $=\frac{n_{1}}{n_{1}+n_{2}} =\frac{0.25}{0.25+5}=0.047$

Given, Mole fraction of NaOH $=\frac{1}{x}$

Then x = 1/mole fraction

⇒   $x=\frac{1}{0.047}$

∴    $x=21.28$

Therefore, the value of x is 21.28.