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 [tex]\frac{weight}{volume}[/tex] in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be [tex]\frac{1000mL}{200mL}=\frac{1g}{xg}[/tex]
200mL × 1g = 1000 mL × x(g)
x(g) = [tex]\frac{200mL*1g}{1000mL}[/tex]
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.
∴ [tex]\frac{10mL}{250mL}=\frac{0.2g}{y(g)}[/tex]
y(g) = [tex]\frac{250mL*0.2g}{10mL}[/tex]
y(g) = 5g of benzalkonium chloride.
Now, at 17% [tex]\frac{weight}{volume}[/tex] 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;
= [tex]\frac{17g}{5g} = \frac{100mL}{z(mL)}[/tex]
z(mL) = [tex]\frac{100mL*5g}{17g}[/tex]
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
