问题 选择题

下图是几种盐的溶解度曲线,根据曲线判断下列说法正确的是:

A.20℃时,硝酸钾溶液的质量分数约为24%;

B.30℃时,将30g硝酸钾和30g氯化钠同时溶于100g水中,蒸发时先析出氯化钠晶体;

C.40℃时,20g食盐溶于100g水中,降温至0℃时可析出晶体;

D.60℃时,将140硫酸铜的饱和溶液降温至30℃,可析出晶体15g。

答案

答案:B

题目分析:由图可知,20℃时硝酸钾的溶解度为31.6g,那么硝酸钾的饱和溶液的溶质质量分数为

31.6g/131.6g ×100%=24%,A错误;30℃时,硝酸钾的溶解度大于氯化钠的溶解度,并且氯化钠的溶解度随温度的变化变化不大,溶解度越小晶体越先析出,故氯化钠先析出,B正确;0℃时,氯化钠的溶解度大于20g,故降温至0℃时没有晶体析出,C错误;60℃时,的溶解度为40g,即100g水中能溶解40g硫酸铜,30℃时硫酸铜的溶解度为25g,由于硫酸铜从溶液中析出为硫酸铜晶体,含有结晶水,故析出晶体的质量大于15g,D错误。故选B。

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Dear Sir,

I’m Li Hua from Fujian.                                           

                                                                               

                                                                              

                                                                                  

                                                                                  

                                                                                     

                                                                                    

                                                                                  

                                                                                       

                                                                                    

                                                                                  

                                                                                   

                                                                                        

                                                                                     

                                                                                          

                                                                                       

                                                                                    

                                                                                  

I am looking forward to your early reply.

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Li Hua