【答案】
分析:A、向溶液中加水,溶質(zhì)質(zhì)量不變,溶質(zhì)的質(zhì)量分數(shù)減小,利用溶質(zhì)質(zhì)量分數(shù)計算公式,計算水的質(zhì)量增加一倍時溶質(zhì)的質(zhì)量分數(shù);
B、向溶液中加水,溶質(zhì)質(zhì)量不變,溶質(zhì)的質(zhì)量分數(shù)減小,利用溶質(zhì)質(zhì)量分數(shù)計算公式,計算增加水的質(zhì)量使溶液質(zhì)量變成2W時溶質(zhì)的質(zhì)量分數(shù);
C、蒸發(fā)水,溶劑質(zhì)量減少而使溶液的溶質(zhì)質(zhì)量分數(shù)增大,至析出固體
克成為飽和溶液,溶質(zhì)的質(zhì)量不再隨水蒸發(fā)而改變;
D、兩種同種溶質(zhì)不同質(zhì)量分數(shù)的溶液混合,混合溶液中溶質(zhì)為兩溶液中溶質(zhì)質(zhì)量和、溶液質(zhì)量為兩溶液的質(zhì)量和,利用溶質(zhì)的質(zhì)量分數(shù)計算公式,計算混合溶液的溶質(zhì)質(zhì)量分數(shù).
解答:解:A、加入相當(dāng)原溶液中溶劑量的水,原溶液中水的質(zhì)量=W-W×20%=0.8W;
加入0.8Wg水后溶液的溶質(zhì)質(zhì)量分數(shù)=
≈11.1%≠10%.故不選A;
B、加水使溶液的質(zhì)量為2W克后溶液的溶質(zhì)質(zhì)量分數(shù)=
=10%.故可選B;
C、蒸發(fā)水,析出
克固體食鹽;蒸發(fā)水溶液質(zhì)量減小而溶質(zhì)質(zhì)量分數(shù)變大,至析出固體達到飽和;繼續(xù)蒸發(fā)水開始析出固體食鹽,但此時溶液的溶質(zhì)質(zhì)量分數(shù)不再改變.因此,不能使溶液的質(zhì)量分數(shù)變成10%,故不選C;
D、加入2W克5%的食鹽溶液后,所得混合溶液的溶質(zhì)質(zhì)量分數(shù)=
=10%.故可選D;
故選BD.
點評:溶液中溶劑的質(zhì)量增加一倍時,溶液的溶質(zhì)質(zhì)量分數(shù)卻不會減小為原質(zhì)量分數(shù)的一半.