You are given a solution that is 518 mm lactose. you need to make up 4.5 l of 16.7 mm solution. what volume do you need to transfer from your stock?

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

Vi = 0.145 l or 145 ml

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

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answer: mole of the gas

answer: mole of the gasexplanation: