Question

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL

Answer 1

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the $\frac{weight}{volume}$ in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be $\frac{1000mL}{200mL}=\frac{1g}{xg}$

200mL × 1g = 1000 mL × x(g)

x(g) = $\frac{200mL*1g}{1000mL}$

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ $\frac{10mL}{250mL}=\frac{0.2g}{y(g)}$

y(g) = $\frac{250mL*0.2g}{10mL}$

y(g) = 5g of benzalkonium chloride.

Now, at 17% $\frac{weight}{volume}$ concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= $\frac{17g}{5g} = \frac{100mL}{z(mL)}$

z(mL) = $\frac{100mL*5g}{17g}$

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride