現(xiàn)有50g質(zhì)量分數(shù)為10%食鹽溶液,要使其質(zhì)量分數(shù)增大1倍,可采用的方法是
A.蒸發(fā)掉一半的溶劑 B.蒸發(fā)掉25g溶劑 C.加入6.5g食鹽 D、加入50g10%NaCl溶液.
【答案】
分析:要使其10%食鹽溶液質(zhì)量分數(shù)增大1倍,即變成質(zhì)量分數(shù)為20%的食鹽溶液,使用溶質(zhì)的質(zhì)量分數(shù)=
,計算所提供四種方法所得溶液的溶質(zhì)質(zhì)量分數(shù),質(zhì)量分數(shù)為20%的方法即為可采用的方法.
解答:解:A、原溶液中溶劑的質(zhì)量=50g-50g×10%=45g,蒸發(fā)掉一半的溶劑,即蒸發(fā)22.5g溶劑,溶質(zhì)質(zhì)量不變,溶液質(zhì)量減小變?yōu)?0g-22.5g=27.5g;此時溶液的溶質(zhì)質(zhì)量分數(shù)=
×100%=18.2%≠20%;故A不正確;
B、蒸發(fā)掉25g溶劑,溶質(zhì)質(zhì)量不變,溶液質(zhì)量減小為50g-25g=25g,此時溶液的溶質(zhì)質(zhì)量分數(shù)=
×100%=20%;故B正確;
C、加入6.5g食鹽,溶質(zhì)質(zhì)量增加,溶液質(zhì)量增加,此時溶液的溶質(zhì)質(zhì)量分數(shù)=
×100%=20%;故C正確;
D、加入50g10%NaCl溶液,所得溶液中溶質(zhì)為兩溶液中溶質(zhì)質(zhì)量之和,溶液為兩溶液質(zhì)量和;
則所得溶液的溶質(zhì)質(zhì)量分數(shù)=
×100%=10%≠20%;故D不正確;
故選BC.
點評:溶質(zhì)質(zhì)量分數(shù)相等的溶液混合后所得溶液的質(zhì)量分數(shù)仍為原溶液的溶質(zhì)質(zhì)量分數(shù),這也是溶液具有均一性的體現(xiàn).