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ANEK [815]
3 years ago
15

What does it mean to standardize the naoh solution why can the molarity of the naoh solution not be determined accurately?

Chemistry
2 answers:
ruslelena [56]3 years ago
5 0
A standard solution is a solution (in this case sodium hydroxide) whose concentration (molarity) is known very precisely. <span>The molarity of the sodium hydroxide solution cannot be determined accurately because s</span>olid sodium hydroxide is highly hygroscopic (absorbs water from the air) and cannot be accurately weighed. Sodium hydroxide form sodium carbonate because it absorbs carbon dioxide from the air.
zhannawk [14.2K]3 years ago
5 0

Answer:

- To set the actual or exact concentration of sodium hydroxide.

- It is hygroscopic.

Explanation:

Hello,

Standardization is a procedure in experimental chemistry to know the actual concentration of a solution, in this case, sodium hydroxide. Such process is carried out by titrating the solution of sodium hydroxide with acid potassium phthalate.

On the other hand, the molarity of sodium hydroxide is not determined accurately since it is a highly hygroscopic material, it means that it includes the water in its structure, therefore the determined molarity will not be accurate since there will be an amount of water included into the sodium hydroxide's exact amount.

Best regards.

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5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

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