求教高三化学题目

commissioner

in order to standardise an approximately 0.2M solution of hydrochloric acid,an analyst first prepare an accurately known solution of sodium carbonate .to do this 5.404g of pure anhydrous sodium carbonate was weighed out,dissolved in water and made up to 500ml on a volumetric flask.Calculate the malor concentratiin of
this standard solution。

commissioner

为了对大约 0.2M 的盐酸溶液进行标准化，分析人员首先制备了准确已知的碳酸钠溶液。为此，称出 5.404g 纯无水碳酸钠，溶于水中，并在体积上定容至 500ml，烧瓶。计算
这个标准溶液

commissioner

看不懂什么意思

david72s

高三化学，这需要多专业的人才能解答啊！

猪胖胖

楼主啊，能读懂你的中文的人都少啊，况且英文，惭愧惭愧。
我看你不如找邻居家理工科大学生问问吧

liangweilwlw

应该是摩尔浓度的意思吧，那就是5.404/（23*2+12+16*3）/0.5=0.102mol/l

liangweilwlw

各元素的摩尔质量我取的整数，可以用题目提供的更精确的，算下来应该大概是0.1mol/l才对

shurman

这个题目其实很简单啊。

yuanlin1988

前半段盐酸都是废话，正题是5.404g无水碳酸钠溶解在500ml水里面，求摩尔浓度。
算出碳酸钠的摩尔质量，然后求出一共多少摩尔，然后除以体积得出浓度。So easy

commissioner

谢谢。我就是不明白第一句0.2M有什么用。其它和上面解答一样

edoowrub

'第一句0.2M'是用来检测学生的思路的.  
换句话说"..........."

zl5431

话说高三化学这么简单吗，我看过物理题还多少有点广度和深度的。

红树林

Insufficient information to answer question, we need to know the volume of HCl added in order to find its concentration. If we knew the volume added use c=n/v where we know the number of moles is 2*moles of sodium carbonate to find concentration. 我家化学满分同学的回答

All_Black

红树林 发表于 2023-1-15 16:13
Insufficient information to answer question, we need to know the volume of HCl added in order to fin ...

感谢学神回复！

Kevinant

红树林 发表于 2023-1-15 16:13
Insufficient information to answer question, we need to know the volume of HCl added in order to fin ...

我不覺得是沒有足夠的資訊,

in order to standardise an approximately 0.2M solution of hydrochloric acid,an analyst first prepare an accurately known solution of sodium carbonate .to do this 5.404g of pure anhydrous sodium carbonate was weighed out,dissolved in water and made up to 500ml on a volumetric flask.Calculate the malor concentratiin of this standard solution。


題目說是要計算這個溶液的濃度

先計算摩爾數 5.404g/ 105.9888g/Mol  (NaCO3的原子量)  =  0.0509865193 Moles

濃度  = 0.0509865193 Moles / 0.5 L = 0.10197 Mol/L

過了16年了還記得一些

shurman

Kevinant 发表于 2023-1-15 18:03
我不覺得是沒有足夠的資訊,

in order to standardise an approximately 0.2M solution of hydrochloric  ...

前面那个0.2M是 用来迷惑学生的。

红树林

Kevinant 发表于 2023-1-15 18:03
我不覺得是沒有足夠的資訊,

in order to standardise an approximately 0.2M solution of hydrochloric  ...

If the question is asking for standard concentration of Na2CO3 then the answer above is correct. If the question was a bit more complex and asked for the concentration of HCl then there is insufficient information, hence perhaps confusion from the asker. The question is not worded very well.
我只是二传手：）

Purplelight

红树林 发表于 2023-1-15 19:03
If the question is asking for standard concentration of Na2CO3 then the answer above is correct. I ...

题目问的是标准溶液. 显而易见 指的是碳酸钠. 因为另一个是被滴定(被测)溶液.

说实话 楼主问这个问题 就像问 1+1 等于几. 如果是一楼 我肯定不敢回答：等于2

Kevinant

红树林 发表于 2023-1-15 19:03
If the question is asking for standard concentration of Na2CO3 then the answer above is correct. I ...

obviously the standard concentration is referring to Na2Co3. The statement about HCL is to give a context.

It's similar to -

There are 6 person lining up outside the restaurant waiting to be sited.

There are 8 seats around the table. How many ways are you able to sit 6 people plus another VIP around the circular table?

冷色调

碳酸钠溶于水生成碳酸氢钠
Na2CO3 + H2O + CO2 = 2 NaHCO3

碳酸氢钠和盐酸中和
NaHCO3 + HCl = NaCl + H2O + CO2

Na2CO3分子量106，NaHCO3分子量84，NaCl 分子量59
Na2CO3称重5.404克，对应产生NaHCO3 8.565克
中和后，消耗HCl 3.773克，相应产生NaCl  6.0156克
原来的盐酸摩尔浓度为0.2，得出原体积18.86升，加上0.5升的碳酸氢钠溶液，中和后体积19.36升，所以标准溶液浓度是6.0156/19.36=0.31 mol/L