In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:

$n_{acid}=M_{base}V_{base}$

Thus, by solving for the volume of the base as NaOH, we obtain:

$M_{base}=\frac{n_{acid}}{V_{base}} \\\\M_{base}=\frac{0.8675g*\frac{1mol}{204.22g} }{0.02456L} \\\\M_{base}=0.173M$

Best regards!

6 0

3 years ago

Estimate the molar mass of a gas that effuses at 1.6 times the effusion rate of CO2

Vlada [557]

To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:

r1/r2 = √(m2/m1)

where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol)

r1 = 1.6r2

1.6 = √(44.01 / m1)
m1 = 17.19 g/mol

7 0

3 years ago