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

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

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be

200mL × 1g = 1000 mL × x(g)

x(g) =

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.

∴

y(g) =

y(g) = 5g of benzalkonium chloride.

Now, at 17% 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;

=

z(mL) =

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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

gas

explanation:

smoke is the matter of a gas