[ss99]

The question goes like "knowing that commercial HNO3 is not a primary standard substance (density 1.42 g/ml, weigh 70%) whereas pure Na2CO3 is a primary standard
Describe how can you prepare 1L of 0.1M HNO3 using both of substance above"
I started by calculating the mass of HNO3 then I didn't know where to go, need some help:(

[chenbeier]

Calculate how much gramm contain 0,1 M HNO₃.

This mass has to be obtain from the given nitric acid. Convert to  the 70%
.
Change it to the volume by using the density.

Then design a experiment by using the carbonate to verify the content.
Now give your ideas.

[ss99]

Then I can say 0.1mol/L*1L*63g/mol*0.7=4.41g which will be the mass of HNO3 that we will use
Then we can say the volume that we will need will be  4.41g÷(1.42g/ml)= 3.10ml of HNO3

Then the moles from standard HNO3 will be equal to the moles of Na2CO3
So we can obtain the mass of Na2CO3 0.1mol*106g= 10.6g of Na2CO3

[chenbeier]

The calculation with  the percentage is wrong. You have to divide by 0.7. The mass obtained with the mole is 100%. So the mass has to be higher if it is only 70%

The sodium carbonate neutralize 2 time of nitric. Why?

[AWK]

Then I can say 0.1mol/L*1L*63g/mol*0.7=4.41g which will be the mass of HNO3 that we will use
Then we can say the volume that we will need will be  4.41g÷(1.42g/ml)= 3.10ml of HNO3

For 0.1 M solution, you need 6.3 g of 100 % nitric acid!
Concentrated 70% acid will weigh a little more.

[ss99]

Then I can say 0.1mol/L*1L*63g/mol*0.7=4.41g which will be the mass of HNO3 that we will use
Then we can say the volume that we will need will be  4.41g÷(1.42g/ml)= 3.10ml of HNO3

For 0.1 M solution, you need 6.3 g of 100 % nitric acid!
Concentrated 70% acid will weigh a little more.

1.42g/ml*1000*(70/100)*(1/63g/mol)
= 15mol/L HNO3 =15.7M this will be the concentration of the 70% acid

(0.1M*1L)/15.7M = 0.00637 L

So to prepare the primary solution we take (0.00637L) of the 15.7M acid and dilue it to 1L

Is this correct?

[chenbeier]

Yes correct, if you had read my answer above and correct your calculation (division instead of multiplikation) then you would have the result more earlier.

[AWK]

= 15mol/L HNO3 =15.7M

Printing and rounding error 15 or 15.7? I got 15.775 M
In the lab you can measure volume to 0.1 ml - I got 6.338, after abbreviation 6.3 ml.

[ss99]

= 15mol/L HNO3 =15.7M

Printing and rounding error 15 or 15.7? I got 15.775 M
In the lab you can measure volume to 0.1 ml - I got 6.338, after abbreviation 6.3 ml.

yeah it 15.7 I misspelled it 😅

[ss99]

= 15mol/L HNO3 =15.7M

Printing and rounding error 15 or 15.7? I got 15.775 M
In the lab you can measure volume to 0.1 ml - I got 6.338, after abbreviation 6.3 ml.

yeah it 15.7 I misspelled it 😅

it's**