Question

某學(xué)生欲配制50g質(zhì)量分?jǐn)?shù)為7.0%的NaCl溶液，在稱量NaCl的過程中，將砝 碼放在左盤（1g以下使用游碼）．該學(xué)生配制的NaCl溶液的質(zhì)量分?jǐn)?shù)實(shí)際是______．

Answer 1

【答案】分析：利用溶質(zhì)質(zhì)量=溶液質(zhì)量×溶質(zhì)的質(zhì)量分?jǐn)?shù)，計(jì)算出溶質(zhì)質(zhì)量和溶劑質(zhì)量，根據(jù)左盤質(zhì)量=右盤質(zhì)量+游碼質(zhì)量，列出等式算出實(shí)際稱量的質(zhì)量，再計(jì)算出溶質(zhì)質(zhì)量分?jǐn)?shù)即可．
解答：解：配制50g質(zhì)量分?jǐn)?shù)為7.0%的NaCl溶液所需溶質(zhì)質(zhì)量為：50g×7.0%=3.5g，需水的質(zhì)量為：50g-3.5g=46.5g，稱量需使用3g的砝碼，將游碼撥到0.5g刻度處，由左盤質(zhì)量=右盤質(zhì)量+游碼質(zhì)量，則3g=氯化鈉的質(zhì)量+0.5g，可得實(shí)際稱得氯化鈉的質(zhì)量為2.5g，故該學(xué)生配制的NaCl溶液的質(zhì)量分?jǐn)?shù)實(shí)際是100%=5.1%．
故答案為：5.1%．
點(diǎn)評：本題難度不大，掌握溶質(zhì)質(zhì)量分?jǐn)?shù)的有關(guān)計(jì)算、左盤質(zhì)量=右盤質(zhì)量+游碼質(zhì)量是解答本題的關(guān)鍵．