Question

（2013?南沙區(qū)一模）把12g某固體放入90g水中充分溶解后過濾，濾出該固體2g，則該濾液的溶質(zhì)質(zhì)量分數(shù)是（　�。�

A．10%

B．11.7%

C．12%

D．13.3%

Answer 1

分析：根據(jù)題意，結(jié)合溶質(zhì)質(zhì)量分數(shù)=

溶質(zhì)質(zhì)量

溶液質(zhì)量

×100%，進行分析計算即可．

解答：解：根據(jù)題意，把12g某固體放入90g水中充分溶解后過濾，濾出該固體2g，即所得溶液中溶質(zhì)的質(zhì)量為12g-2g=10g，所得溶液的質(zhì)量為10g+90g=100g；
濾液的溶質(zhì)質(zhì)量分數(shù)=

溶質(zhì)質(zhì)量

溶液質(zhì)量

×100%=

10g

100g

×100%=10%．
故選A．

點評：本題很簡單，掌握溶質(zhì)質(zhì)量分數(shù)的計算方法是正確解答本題的關(guān)鍵．