Word problems please help

[mkay]

A nurse wants to add water to 30 oz. of a 10% solution of benzalkonium chloride to dilute it to an 8% solution. How much water must she add? (hint: water is 0% benzalkonium chloride.)

The 0% has thrown me off.

 

[Deleted member 4993]

A nurse wants to add water to 30 oz. of a 10% solution of benzalkonium chloride to dilute it to an 8% solution. How much water must she add? (hint: water is 0% benzalkonium chloride.)

The 0% has thrown me off.

That means water does not have any benzalkonium chloride → when you are adding water to the solution - the amount of benzalkonium chloride remains unchanged.

 

[mkay]

Still stuck

The 8% solution needed I think would be (.08 x 30)= 2.4
The solution of benz. chloride would be .10(x + 30)x being the water
It's adding the water in I can't figure out how to do.

 

[Mrspi]

A nurse wants to add water to 30 oz. of a 10% solution of benzalkonium chloride to dilute it to an 8% solution. How much water must she add? (hint: water is 0% benzalkonium chloride.)

The 0% has thrown me off.

benzalkonium chloride in original + benzalkonium chloride added = benzalkonium chloride at the end

You start with 30 oz of 10% solution....how much benzalkonium chloride is in that? 10% of 30 oz, right? or 0.10(30) oz

You are adding WATER which has 0% benzalkonium chloride in it. Let x = number of ounces of water you add. How much bc are you adding? Well, you are adding 0.00(x) ounces.

After you add x ounces of water, you will have (30 + x) ounces of solution, which is 8% benzalkonium chloride. So how much bc do you have at the END of the mixing? 0.08(30 + 8)

.10(30) + 0(x) = 0.08(30 + x)
3 = 0.08(30 + x)

Solve that for x.....