Question

In a volume of 0.400 dm^3 of CH2CL2 there are 3.73*10^24 molecules at 25 degree celsius. Calculate
(A) The density of the substance at 25 degree celsius, expressed in g.cm^3.
(B) The number of atoms present in 425g of CH2CL2. (Solve by means of complete numerical development without omitting the statements or the units)

Answer 1

Answer on Question #65158 - Chemistry - Organic Chemistry

Task:

In a volume of 0.400 dm³ of CH₂Cl₂ there are 3.73*10²⁴ molecules at 25 degree celsius.

Calculate:

(A) The density of the substance at 25 degree celsius, expressed in g.cm³.

(B) The number of atoms present in 425g of CH₂Cl₂. (Solve by means of complete numerical development without omitting the statements or the units)

Solution:

Part (A):

We find the amount of CH₂Cl₂, using the following formula:

$$n = \frac {N}{N _ {a}};$$ n \left(\mathrm{CH_2Cl_2}\right) = \frac {N \left(\mathrm{CH_2Cl_2}\right)}{N _ {a}} = \frac {3.73 \times 10^{24}}{6.022 \times 10^{23}} = 6.194 \text{ (moles of } \mathrm{CH_2Cl_2\text{)}

We find the mass of CH₂Cl₂, using the following formula:

$$n = \frac {m}{M}; \quad \Rightarrow \quad m = n \times M$$ $$M \left(\mathrm{CH_2Cl_2}\right) = 84.93 \frac{g}{mol}$$ $$m \left(\mathrm{CH_2Cl_2}\right) = n \left(\mathrm{CH_2Cl_2}\right) \times M \left(\mathrm{CH_2Cl_2}\right);$$ $$m \left(\mathrm{CH_2Cl_2}\right) = 6.194 \text{ mol} \times 84.93 \frac{g}{mol} = 526.056 \text{ g}$$

Convert dm³ in L:

$$1 \text{ dm}^3 = 1 \text{ L} = 1000 \text{ mL} = 1000 \text{ cm}^3;$$ $$0.400 \text{ dm}^3 = X \text{ mL};$$ $$X = V \left(\mathrm{CH_2Cl_2}\right) = \frac {1000 \text{ cm}^3 \times 0.400 \text{ dm}^3}{1 \text{ dm}^3} = 400 \text{ cm}^3$$

We find the density of the substance, using the following formula:

$$\rho = \frac {m}{V};$$ $$\rho \left(\mathrm{CH_2Cl_2}\right) = \frac {m \left(\mathrm{CH_2Cl_2}\right)}{V \left(\mathrm{CH_2Cl_2}\right)} = \frac {526.056 \text{ g}}{400 \text{ cm}^3} = 1.31514 \frac{g}{\text{cm}^3}$$

Answer (A): The density of the substance is $1.31514\ \mathrm{g}\cdot \mathrm{cm}^{-3}$

Part (B):

We find the amount of $\mathrm{CH}_2\mathrm{Cl}_2$, using the following formula:

$$n = \frac{m}{M};$$ $$n(\mathrm{CH}_2\mathrm{Cl}_2) = \frac{m(\mathrm{CH}_2\mathrm{Cl}_2)}{M(\mathrm{CH}_2\mathrm{Cl}_2)} = \frac{425\ \mathrm{g}}{84.93\ \mathrm{g/mol}} = 5.004\ (\mathrm{moles\ of\ CH_2Cl_2})$$

We find the number of molecules of $\mathrm{CH}_2\mathrm{Cl}_2$, using the following formula:

$$n = \frac{N}{N_a}; \quad \Rightarrow \quad N = n \times N_a;$$ $$N(\text{molecules}) = n(\mathrm{CH}_2\mathrm{Cl}_2) \times N_a;$$ $$N(\text{molecules\ of\ CH}_2\mathrm{Cl}_2) = 5.004 \times 6.022 \times 10^{23} = 30.134 \times 10^{23}$$

In molecule $\mathrm{CH}_2\mathrm{Cl}_2$ contains 5 atoms. Then,

$$N(\text{atoms}) = 5 \times N;$$ $$N(\text{atoms\ of\ CH}_2\mathrm{Cl}_2) = 5 \times 30.134 \times 10^{23} = 1.5067 \times 10^{25}$$

Answer (B): The number of atoms $= 1.5067 \cdot 10^{25}$.

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