Answer:
Volume of acid, Va=250mL; Volume of quinine,Vb=20mL; Molarity of acid, Ma=0.05M.
Molar mass of acid= H2+S+O4= 2+32+(16X4)= 2+32+64=98g
Concentration of acid, Ca= Molar mass of acid/ Ma =98/0.05=1960g/mol
Explanation: To calculate concentration of quinine, Cb is as follow
Va*Ca=Vb*Cb
∴ Cb=Va*Ca/Vb =250*1960/20 =24500g/mol
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M