今有60g溶質(zhì)的質(zhì)量分?jǐn)?shù)為10%的NaNO3溶液,欲將其溶質(zhì)的質(zhì)量分?jǐn)?shù)增大一倍,應(yīng)采用的方法是( )
A.把溶劑蒸發(fā)掉一半
B.加入6gNaNO3晶體
C.把溶劑蒸發(fā)掉30g
D.加入20%的NaNO3溶液30g
【答案】
分析:欲將60g溶質(zhì)的質(zhì)量分?jǐn)?shù)為10%的NaNO
3溶液溶質(zhì)的質(zhì)量分?jǐn)?shù)增大一倍,即通過改變?nèi)芤旱慕M成使溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)變?yōu)?0%的溶液;分析所采用的方法對溶液組成的改變,利用溶質(zhì)質(zhì)量分?jǐn)?shù)的計算式求出改變后溶液的溶質(zhì)質(zhì)量分?jǐn)?shù),做出判斷.
解答:解:A、把溶劑蒸發(fā)掉一半,則溶液的質(zhì)量=60g-60g×(1-10%)×
=35g,所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=
×100%≈17%;故A不正確;
B、加入6gNaNO
3晶體,所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=
×100%≈18.2%;故B不正確;
C、把溶劑蒸發(fā)掉30g,則溶液的質(zhì)量=60g-30g=30g,所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=
×100%=20%;或蒸發(fā)溶劑質(zhì)量為原溶液質(zhì)量一半時,由于溶質(zhì)質(zhì)量不變而溶液質(zhì)量減小一半,因此溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)增大一倍;故C正確;
D、加入20%的NaNO
3溶液30g,則所得混合溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)=
×100%≈13%;故D不正確;
故選C.
點評:采取蒸發(fā)溶液質(zhì)量一半的方法可使溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)增大一倍;采取使原溶液中溶質(zhì)質(zhì)量增大一倍,由于溶液質(zhì)量也隨之增加,而不能使溶液質(zhì)量分?jǐn)?shù)增大一倍.