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Aliun [14]
4 years ago
10

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1? 6.4 × 10−4 g 6.4 × 10−3 g

6.4 × 10−2 g 6.4 × 10−1 g
Chemistry
1 answer:
eduard4 years ago
8 0

Compete Question:

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?

               Passage: "16 mmol of CDP in 1 L of buffer"

                                       

Answer:

6.4 × 10-2 g  

Explanation:

  Mass = Mole × Molar Mass

   we are given from the question that 16 mmol of CDP is in 1 L of buffer

    this mean that we have 16 × 10^-3 moles of CDP in 1 liter of buffer.

so the mass of CDP in one liter of buffer will be calculate as,

         mass of CDP = 16 × 10^-3 × 403g mol−1

                               = 64 × 10^-1

                               = 6.4 g/L

But because the question

asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:

6.4 g/L × 10 mL

6.4 g/L × 0.01 L  = 6.4 × 10^-2

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