According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m
Answer:
<h3>The answer is 7.42 </h3>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H_3O}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H_3O%7D%5E%7B%2B%7D%5D)
From the question we have
We have the final answer as
<h3>7.42 </h3>
Hope this helps you
The concentration after dilution is 1.4%.
We are aware that concentration and volume are related to each other by the formula -
= 
, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.
Keep the values to calculate final concentration.
= (53.5 × 5.4)/205.0
Performing multiplication on Right and Side
= 288.9/205.0
Performing division on Right Hand Side
= 1.4%
Hence, the final concentration is 1.4%.
Learn more about concentration -
brainly.com/question/17206790
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The complete question is -
A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.
Calculate the final concentration and express your answer to two significant figures and include the appropriate units.
Valence electrons are the electrons in the outermost shell of an element on the periodic table. Atoms want to be able to have a full outer shell and they can share or trade electrons in order to achieve this. Valence electrons are also super super important in chemical reactions. The number of valence electrons determines what group that specific atom or element is in on the periodic table. This affects the reactivity of the element.
