Question

某醫(yī)院需配制溶質(zhì)質(zhì)量分?jǐn)?shù)為0.9%的生理鹽水1000g，需將多少克醫(yī)用氯化鈉溶在多少毫升蒸餾水中？（蒸餾水的密度為1g/cm³）

Answer 1

【答案】分析：已知溶液質(zhì)量和溶質(zhì)的質(zhì)量分?jǐn)?shù)，根據(jù)溶質(zhì)質(zhì)量=溶液質(zhì)量×溶質(zhì)的質(zhì)量分?jǐn)?shù)即可求的氯化鈉溶質(zhì)的質(zhì)量；再根據(jù)溶液質(zhì)量=溶質(zhì)質(zhì)量+溶劑質(zhì)量即可求得所需水的質(zhì)量；最后再根據(jù)V=即可求得水的體積．
解答：解：溶液質(zhì)量為1000g，溶質(zhì)質(zhì)量分?jǐn)?shù)為0.9%；
則NaCl的質(zhì)量=1000g×0.9%=9g；
水的質(zhì)量=溶液質(zhì)量-溶質(zhì)質(zhì)量=1000g-9g=991g；
則水的體積==991cm³．
答：需用9g醫(yī)用氯化鈉溶于991mL蒸餾水中．
點(diǎn)評(píng)：本題考查了溶質(zhì)質(zhì)量分?jǐn)?shù)的簡單計(jì)算；解本題的關(guān)鍵就是正確理解溶質(zhì)的質(zhì)量分?jǐn)?shù)這一概念．