要配制0.10mol/L的HCl溶液1000ml,需要密度为1.19g/ cm³、质量分数0.37 要配制0.10摩尔每升的HCL溶液1000mL需密度为1.1...
\u8981\u914d\u52360.1mol/L\u7684HCl\u6eb6\u6db21000ml\uff0c\u9700\u8981\u5bc6\u5ea6\u4e3a1.19g/cm\uff0c\u8d28\u91cf\u5206\u6570\u4e3a0.37\u7684\u6d5384ml\uff0c\u6c42\u91c7\u7eb3
解: 1000ml=1L
HCl的物质的量为:0.1mol/L×1L=0.1mol
HCl的质量为: 0.1mol×36.5g/mol=3.65g
需要质量分数为0.37的浓盐酸的质量为:3.65g÷0.37=9.86g(取小数点后两位小数)
需要浓盐酸的体积为:9.86g÷1.19g/cm3=8.29cm3
所以配制0.10mol/L的HCl溶液1000ml,需要密度为1.19g/ cm³、质量分数0.37的浓盐酸8.29ml
HCL分子量36.5
36.5*0.1*1=1.19*V*0.37
V=8.3
绛旓細瑕侀厤鍒0.10鎽╁皵姣忓崌鐨凥Cl婧舵恫锛岄渶瑕佺煡閬撴墍闇鐨勮川閲忓垎鏁版祿搴﹀拰鎽╁皵璐ㄩ噺銆侶Cl鐨勬懇灏旇川閲忎负36.46 g/mol銆傞鍏堣绠楁墍闇HCl鐨勬懇灏旇川閲忎箻浠ユ墍闇鐨勬懇灏旀祿搴︼紝寰楀埌姣忓崌婧舵恫涓墍闇鐨勮川閲忎负锛0.10 mol/L 脳 36.46 g/mol = 3.646 g/L 鐒跺悗鏍规嵁鎵闇鐨勮川閲忓垎鏁拌绠楁墍闇鐨勫瘑搴︼紝鍗筹細蟻 = (0.37 脳 1...
绛旓細闇瑕佽川閲忓垎鏁颁负0.37鐨勬祿鐩愰吀鐨勮川閲忎负锛3.65g梅0.37=9.86g(鍙栧皬鏁扮偣鍚庝袱浣嶅皬鏁)闇瑕佹祿鐩愰吀鐨勪綋绉负锛9.86g梅1.19g/cm3=8.29cm3 鎵浠閰嶅埗0.10mol/L鐨凥Cl婧舵恫1000ml锛岄渶瑕佸瘑搴︿负1.19g/ cm³銆佽川閲忓垎鏁0.37鐨勬祿鐩愰吀8.29ml
绛旓細鐢遍鎰忓緱锛宮锛圚Cl锛=0.1mol*36.5g/mol=3.65g m锛堟祿鐩愰吀锛=3.65g/37%=365/37g V=锛365/37g锛/1.19g/cm涓夋鏂圭害绛変簬8.2898ml 璇稬Z閲囩撼锛岃阿璋紒
绛旓細c=n/v=(w*p*v)/(M*V)=(0.37*1.19*10^3*v)/(36.5*1)v=8.2898*10^(-3)L=8.2898ml
绛旓細瑕侀厤鍒100mL娴撳害绾︿负0.10mol/L鐨 K2Cr2O7鏍囧噯婧舵恫锛岄渶瑕佹寜鐓т笅闈㈢殑姝ラ鎿嶄綔锛1. 璁$畻鎵闇鐨 K2Cr2O7 鐨勮川閲忥細鎽╁皵璐ㄩ噺(M) = 294.18 g/mol 鎵闇鐗╄川鐨勬懇灏旀暟(n) = 0.10 mol/L 脳 0.100 L = 0.010 mol 鎵闇鐗╄川鐨勮川閲(m) = M 脳 n = 29.418 g 2. 灏嗚绠楀嚭鐨 K2Cr2O7 璐ㄩ噺绉...
绛旓細閰嶅埗0.10 mol/L NaOH 婧舵恫鐨勫疄楠屾楠ゅ涓嬶細鏉愭枡锛歂aOH鍥轰綋绮夋湯 绾按 50 mL瀹归噺鐡 绉伴噺鍣 纾佸姏鎼呮媽鍣ㄦ垨鐜荤拑妫 姝ラ锛氱О鍙4鍏婲aOH鍥轰綋绮夋湯锛屽苟鍔犲叆涓涓共鍑鐨50 mL瀹归噺鐡朵腑銆傚姞鍏ラ傞噺鐨勭函姘达紝鎼呮媽婧惰ВNaOH锛岀洿鑷虫憾娑蹭綋绉帴杩50 mL銆傚皢婧舵恫浣撶Н璋冩暣鑷50 mL锛屽彲浠ヤ娇鐢ㄧ鍔涙悈鎷屽櫒鎴栫幓鐠冩鍏呭垎娣峰悎銆傚皢...
绛旓細L-1鐨勬阿姘у寲閽犵殑婧舵恫鍚湁2.0g姘㈡哀鍖栭挔锛屾墍浠ユ憾娑茬殑浣撶Н涓2.0g40g/mol梅0.10mol/L=0.5L=500mL鎵浠ュ閲忕摱鐨勮鏍煎簲閫夋嫨500ml锛庢搷浣滄楠ゆ湁绉伴噺銆佹憾瑙c佺Щ娑层佹礂娑ゃ佸畾瀹广佹憞鍖绛夋搷浣滐紝涓鑸敤鎵樼洏澶╁钩绉伴噺锛堜娇鐢ㄥ皬鐑ф澂锛夛紝鐢ㄨ嵂鍖欏彇鑽搧锛屽湪鐑ф澂锛堢儳鏉50mL鍗冲彲锛岃繃澶т娇鐢ㄤ笉鏂逛究锛変腑婧惰В锛屽苟鐢ㄧ幓鐠冩...
绛旓細37锛呯殑娴撶洂閰稿瘑搴︿负1.19g/ml,500ml鎵鍚函hcl鐨勮川閲忎负1.19 500 37 = 220.15g 锛屾崲绠楁垚鐗╄川鐨勯噺灏变负220.15 /36.5 = 6.0315mol 閰嶇疆濂界殑婧舵恫閲屾墍鍚殑绾痟cl鐨勭墿璐ㄧ殑閲忔槸涓嶄細鍙樼殑锛屾墍浠閰嶅埗濂界殑绋鐩愰吀鐨勪綋绉氨涓6.0315/2=3.01575l.鎵鍔犲叆鐨勬按灏辨槸3.01575 - 0.5 = 2.51575l ...
绛旓細HCL 闇0.10*1=0.1mol 鍗36.5g/mol*0.1mol=3.65g so v=3.65/1.19 =3.067ml
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