Question

What is the density of 53.4 wt queous naoh solution if 16.7 ml of the solution diluted to 2.00 l gives 0.169 m naoh?

Answer 1

The density of 53.4 wt aqueous NaOH solution is 0.809 g/ml

Given data:

• The mass percent of NaOH is 53.4.
• Volume of NaOH diluted is 16.7 ml.
• The volume of diluted solution is 2.00 L =2000 ml.
• Concentration of diluted solution is 0.169 M.

First, we find the initial concentration of NaOH by using the following formulae,

M₁V₁ = M₂V₂

Where,

M₁ is the initial molarity of NaOH

M₂ is the molarity after dilution

V₁ is the initial volume of NaOH

V₂ is the final volume after dilution.

Substituting the values,

M₁ × 16.7 ml = 0.169 M × 2000 ml,

M₁ =  $\frac{0.169 M *2000 ml}{16.7 ml}$

M₁ = 20.2 M.

Thus, the initial concentration of NaOH is 20.2 M.

we know, 1 M solution contains 1 mol of substance present in 1 L solution,

Thus, 20.2 M solution will have 20.2 mols of NaOH.

Now, we can find the mass of NaOH by using the number of moles and molar mass.

• molar mass of NaOH is 40 g/mol.

Mass = no. of moles × molar mass

= 20.2 mol × 40 g/mol

= 808 g.

Thus, the mass of NaOH is 808g.

53.4 wt of NaOH means 53.4 g of NaOH in a 100 g solution,

Thus, 808 g of NaOH will be present in ,

⇒ $\frac{53.4 g NaOH}{100 g solution} = \frac{808 g NaOH}{x g solution}$

⇒ 1513.1 g

Now, Convert the grams of NaOH to milliliters, using the density of NaOH at room temperature.

• The density of NaOH at room temperature is 1.515 g/ml,

Density = $\frac{mass}{volume}$

⇒ 1.515 g/mol = $\frac{1513.1 g}{volume}$

⇒ volume = $\frac{1513.1 g}{1.515 g/mol}$

⇒ volume = 998.7 ml.

Thus, the volume of NaOH is 998.7 ml.

Hence, we know,

• the mass of NaOH is 808 g
• the volume of NaOH is 998.7 ml

Substituting the values,

Density = 808 g / 998.7 g/ml

⇒ Density = 0.809 g/ml

Thus, the density of 53.4 wt aqueous NaOH is 0.809 g/ml.

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