We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Explanation:
hope it helps you understand moles
1.)b
2.)true
3.)false
are the answer don't take me on my word
Answer:
B) 0.32 %
Explanation:
Given that:

Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)


Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
Answer:
10 times more acidic than pH5 and 100 times more acidic than pH6
Explanation: