Question

what is the total number of newtron in 10 gram of D2O?

Answer 1

Answer on Question 68933, Physics, Atomic and Nuclear Physics

Question:

What is the total number of neutrons in 10 g of $D_2O$?

Solution:

Let's first find the molar mass of the deuterium oxide:

$$M_{D_2O} = M_{D_2} + M_O = 2 \cdot (2\,g \cdot mol^{-1}) + 16\,g \cdot mol^{-1} = 20\,g \cdot mol^{-1}.$$

Then, we can find the number of moles of deuterium oxide molecules in 10 g of $D_2O$:

$$n = \frac{m_{D_2O}}{M_{D_2O}},$$

here, $m_{D_2O}$ is the mass of $D_2O$, $M_{D_2O}$ is the molar mass of the deuterium oxide.

Then, we get:

$$n = \frac{m_{D_2O}}{M_{D_2O}} = \frac{10\,g}{20\,g \cdot mol^{-1}} = 0.5\,mol.$$

Also, we need to find the number of neutrons in one molecule of $D_2O$:

$$N_n = N_n(D_2) + N_n(O) = 2 \cdot 1\,n + 8\,n = 10\,n.$$

Finally, we can find the total number of neutrons in 10 g of $D_2O$:

$$N_{n(total)} = n \cdot N_A \cdot N_n,$$

here, $N_A = 6.02 \cdot 10^{23}\,mol^{-1}$ is the Avogadro constant.

Then, we get:

$$\begin{array}{l} N_{n(total)} = n \cdot N_A \cdot N_n = 0.5\,mol \cdot 6.02 \cdot 10^{23}\,mol^{-1} \cdot 10\,neutrons = \\ = 3.01 \cdot 10^{24}\,neutrons. \end{array}$$

Answer:

$$N_{n(total)} = 3.01 \cdot 10^{24}\,neutrons.$$

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