下表是氯化铵在不同温度下的溶解度：
| 温度（℃） | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
| 溶解度（g/100g水） | 29.4 | 33.3 | 37.2 | 41.4 | 45.8 | 50.4 | 55.0 | 60.2 |
（1）从表中可知：随温度升高氯化铵的溶解度 。
（2）60℃时，向盛有50 g水的烧杯中加入30 g氯化铵，充分搅拌，可得到氯化铵的 （填“饱和”或“不饱和”）溶液，溶质的质量分数是 。
（3）A烧杯是70℃时，溶剂为100g的氯化铵不饱和溶液，经过如下图的变化过程（在整个过程中不考虑水分的蒸发）。
Ⅰ. C烧杯溶液中溶质的质量为 g。
Ⅱ.70℃时，将A烧杯中的溶液恒温蒸发10 g溶剂后，析出a g固体，再继续恒温蒸发30g溶剂，则析出固体b克，则b克 3a 克（填“<、≤、>、≥或=”）。
Ⅲ.有关三个烧杯中溶液的说法，错误的是 （填字母序号）。
a．B烧杯中溶液是不饱和溶液
b．A、B、C烧杯中的溶液在变化过程中，溶解度一定改变
c．由60℃降温至50℃时溶液质量一定减少
d．降温至60℃时，溶液中溶质质量分数一定不变
e．降温至30℃，烧杯中一定有NH4Cl晶体析出
（1）增大
（2）饱和 ，35.5%或11/31
（3）37.2 ，＞ ，a、c、e
题目分析：（1）从表中可知：随温度升高氯化铵的溶解度增大；
（2）60℃时，氯化铵的溶解度为55g，即100g水中可溶解55g达到饱和，现50 g水的烧杯中加入30 g氯化铵，氯化铵只能溶解22.5g，由固体剩余，得到的是饱和溶液，溶质的质量分数为
22.5g／72.5g ×100% =35.5% ；
（3）C烧杯溶液是饱和溶液，20℃氯化铵的溶解度为37.2g，故C烧杯溶液中溶质的质量为37.2g ；A烧杯是70℃时，溶剂为100g的氯化铵不饱和溶液，开始10 g溶剂后，析出a g固体后得到的是饱和溶液，故b克＞3a 克；B烧杯中溶液中无晶体析出，可能是不饱和溶液，也可能恰好饱和；溶解度由温度有关，温度变，则溶解度变；由60℃降温至50℃时若溶液仍是不饱和溶液，则溶液质量不变；溶液中溶质质量分数由溶质和溶剂的质量决定，溶质和溶剂的质量不变，则溶质质量分数一定不变；降温至30℃，若达到饱和，则由晶体析出，若不饱和，则无晶体析出。
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