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3 votes
How many grams of KI are needed to prepare 2,000. grams of an aqueous solution

containing 25 parts per million (ppm) of solute?

User Moho
by
5.4k points

1 Answer

3 votes

Answer:


m_(solute)=0.05g

Step-by-step explanation:

Hello there!

In this case, according to the equation for the calculation of the parts per million:


ppm=(m_(solute))/(m_(solution)) *1x10^6

It is possible to solve for the mass of KI (solute) as shown below:


m_(solute)=(ppm*m_(solution))/(1x10^6)

Thus, we plug in to obtain:


m_(solute)=(25*2,000g)/(1x10^6) \\\\m_(solute)=0.05g

Best regards!

User Ben Stott
by
5.6k points