Question

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

Answer 1

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

Step-by-step 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`