The weight percent formaldehyde in the pesticide sample if 37.49 of 0.2664 were needed to titrate the liberated base 24.218% .
Given mass of sample = 37.49 g
Moles of HCl = ?
Molarity of HCl = 0.2664 M = 0.2664 mol/L
Volume of HCl = 37.49 ml = 37.49*10^-3 L
Moles = Molarity * Volume
Moles of HCl = 0.2664 (mol/L) *37.49*10^-3 L
= 9.987*10^-3 mol .
Neutralization reaction : moles of acid = moles of Base
Moles of Base = 9.987*10^-3 mol .
Moles of Formaldehyde = 9.987*10^-3 mol .
Atomic mass of H = 1 u
O = 16 u
C = 12 u
Molar mass of HCHO = 1+12+1+16 = 30 g/mol .
Moles = mass/molar mass
Mass of Formaldehyde = 9.987*10^-3 mol * 30 g/mol
= 299.6*10^-3 g = = 0.2996 g .
Mass or weight percentage = Mass of Formaldehyde *100/Total mass of sample
Mass % = 0.2664*100/1.10 = 26.64/1.10 = 24.218% .
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