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AlexFokin [52]
3 years ago
7

What is the freezing point in °C) of a 0.195 m

Chemistry
2 answers:
Elina [12.6K]3 years ago
4 0

Answer:

T_S=-1.09\°C

Explanation:

Hello!

In this case, since the freezing point depression for a solution is computed via:

(T_S-T_W)=-imK_f

Whereas TW is the freezing temperature of water, TS that of the solution, i the van't Hoff's factor (3 for K2S as it ionizes properly), m the molality of the solution and Kf the freezing point constant of water. Thus, we plug in to obtain:

(T_S-0\°C)=-3*0.195m*1.86\frac{\°C}{m}\\\\T_S=-1.09\°C

Best regards!

Kay [80]3 years ago
4 0

The correct answer is -1.088. (Don't forget the negative sign).

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<h2>Required Answer :- </h2>

- In the question it is asked how many numbers of protons are present in a sodium atom.

- We know that the atomic number of sodium is 11.

- Sodium atom means it is in its neutral state, the neutral sodium atom has no charge on it.

- We know that the atomic number of an element is always equal to the number of protons for the particular element.

- Therefore the atomic number of sodium is 11 and the number of protons in sodium are 11.

<h3>Additional information :- </h3>

  • If we know the atomic number of an element then only we can find the number of protons, neutrons and electrons present in it. The number of electrons in an element is equal to the atomic number of the respective element

  • The atomic weight of the sodium is 23.

  • The number of neutrons in sodium = Atomic weight of sodium – number of protons in sodium

= 23 – 11 = 12.

  • So, the number of neutrons in sodium is equal to 12.

  • The number of electrons in sodium is equal to the atomic number of sodium. Then the number of electrons in sodium is equal to 11.

  • Therefore the number of protons, neutrons and electrons in sodium are 11, 12 and 11 respectively.

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3 years ago
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