Three teaspoons of sucrose(17.1 g) is dissolved in hot water to make a total volume of 250 ml solution. calculate the molarity

me by showing work

very confused on what to do

Molar mass sucrose :

C12H22O11 =  342.29 g/mol

number of moles of the solute :

n = mass solute / molar mass

n = 17.1 / 342.29

n = 0.05 moles

Volume in liters :

250 mL / 1000 =  0.25 L

therefore :

Molarity  =  moles of solute / volume

Molarity =  0.05 / 0.25  => 0.2 mol/L

Hey there!

The pH range of a buffer is the pH window where in the uffer resists the changes in pH of the solution upon addition of small amounts of acid or base. Generally, the pH range of a buffer solution is equal to pKa ±1.

for this system :

ka= 1.3*10⁻⁷ , hence

pKa= - log (Ka)

pKa =  - log ( 1.3*10⁻⁷)

pKa = 6.88

Therefore, hence the buffer can be used in pH range of 5.88 to 7.88.

Hope that helps!