Could someone help me to solve my chemistry problem?

 

15 ml HNO3 are neutralized by 30 ml KOH.

and beside that, 20ml of this solution of HNO3, after having reacted with 0,700 gr of Na2CO3, are neutralized by 25 ml of the same solution of KOH.

 

Find out molar concentration of both solutions

 

Thank you very much for helping me

 

HNO3 + KOH->KNO3 + H2O

 

2HNO3 + Na2CO3->2NaNO3+ CO2+H2O

 

 

 

 

"Question: Find out the molar concentration of acetic acid (CH₃COOH) solution when 1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution?

 

 

 

 

 

Solution:

Given
mass of acetic acid (w) = 1.24 g

volume of acetic acid (V) = 155.0 ml

= (155.0 / 1000) L {Because 1 Liter = 1000 ml}

= 0.155 L

Molar mass of acetic acid (W) = atomic mass of C atom + atomic mass of 2 O atoms + atomic mass of 4 H atoms

= (12.0 + 2 * 16.00 + 4 * 1.0) g / mol
= 48.0 g / mol

And now by the help of molar concentration (molarity) formula we can determine molar concentration of acetic acid,

Molar concentration of acetic acid (M) = mass of acetic acid (w) / molar mass of acetic acid (W) * volume of acetic acid (V)

= 1.24 g / (48.0 g / mol * 0.155 L)

= 0.167 M

So when 1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution, molar concentration of this solution is 0.167 M."

http://www.tutorvista.com/chemistry/molar-concentration-formula

 

 

Yeah, this is copied, but it should help you a little. I DID NOT know ANY of this stuff, this is from the webpage.