Answer:
0.05 mL
Explanation:
Initially, we have a concentrated solution of NaOH, to which we will add water to get a dilute one. To calculate the volume of the initial solution that we have to measure, we can use the dilution rule:
C₁ × V₁ = C₂ × V₂
where,
C₁ is the initial concentration (50% w/w; ρ = 1.52 g/mL)
V₁ is the initial volume
C₂ is the final concentration (0.1 M)
V₂ is the final volume (10 mL)
First, we have to calculate the weight/volume percentage and then the molarity of the first solution.
[tex]\% w/v = \% w/w \times \rho = 50 g\% g \times 1.52 g/mL = 76g\% mL[/tex]
[tex]M=\frac{m}{M \times V(L) } = \frac{76g}{40g/mol \times 0.1 L} = 19 M[/tex]
Now, we can apply the dilution rule.
[tex]V_{1}=\frac{C_{2} \times V_{2} }{C_{1}} =\frac{0.1M \times 10 mL }{19 M} =0.05mL[/tex]
The volume of solution required is 0.05 mL.
We are told that the solution is a 50 % (w/w) sodium hydroxide solution. This means that the mass of NaOH in the solution is 50g.
First we must obtain the %w/v as follows;
%w/w × density of solution = 50 % × 1.52 g/mL = 76%
To obtain the molarity of the original solution;
76/40 g/mol × 0.1 M
19 M.
Using the dilution formula;
C1V1 = C2V2
C1 = 19 M
V1 = ?
C2 = 0.1 M
V2 = 10 mL
V1 = 0.1 M × 10 mL/19 M = 0.05 mL
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