就是每千克溶剂的溶质摩尔数。
我想罗嗦一下，质量，而不是摩尔数。
对n的偏导数，这里的n是摩尔数。
显然是质量守恒，而不是摩尔数守恒。
然后是A的总质量,除以A的摩尔数。
Then I have the total mass of A divided by the number of moles of A.
平衡时让我们写出摩尔数。
减小压强的方法就是，减小容器内的摩尔数。
And the way to decrease the pressure is to decrease the number of moles in the container.
气体和液体中总的摩尔数之比。
乘以总的气体摩尔数。
即压强乘以体积等于，摩尔数乘以。
This is the pressure times the volume RT equals the number of moles times RT.
只有溶剂的摩尔数。
所以它是摩尔数乘以。
它告诉了你摩尔数。
生成物的摩尔数减去反应物的摩尔数。
因为你在那得到摩尔数，如果你问热容有多大？
Because you've got a factor of the number of moles in there. If you ask how big is the heat capacity?
或许会有特殊的例子,摩尔数在前后是一样的。
So it may be that there's a special case where the number of moles before and after are the same.
我想到摩尔数少一点的地方。
每千克溶剂的溶质摩尔数。
我得到了，气相和液相的，摩尔数的比的表达式。
And I derived an expression for the ratio of the number of moles in the gas and liquid phases.
怎样减小摩尔数？
是已经反应的摩尔数，除以初始摩尔数。
It's the number of moles that have reacted divided by number of moles initially.
除以开始时的摩尔数，也就是n也就是我们想要的。
Divided by the number of moles initially, which is n. That's what we want.
并把总摩尔数拿出来。
对的，它只是摩尔数，乘以每种情况的化学势。
Right, it's just a number of moles times the chemical potential in each case.
摩尔数保持不变，因为是闭系，气体不会跑出去。
The number of moles stays the same a closed systems, gas doesn't come out.
我们想知道已经反应的摩尔数，这是已经反应掉的摩尔数。
X We want the ratio of the number of moles reacted, which is x, that's the number of moles that's gone.
所以我们将气体的摩尔数，改变了3摩尔，好，它会起多大作用？
Here just two, so we changed the number of moles of gas by three. All right, how much did it matter, right?
温度的变化量，正比于b的摩尔数，比例常数和混合物是什么有关。
Amount of temperature change depends linearly on the molality of B, with a constant that depends on what the mixture is.
然后我们需要知道，总摩尔数，所以我们要计算，摩尔比摩尔分数。
And then we're going to need the total number of moles, because we're going to be doing mole ratios, mole fractions.
一种溶质的摩尔浓度，通常被表示作每，000克溶剂的摩尔数。
The molal concentration of a solute, usually expressed as the number of moles of solute per, 000 grams of solvent.
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