Respuesta :
The total number of moles of solute particles present in 1L (exact) of aqueous 0.057 m hno3 is 0.166 mol
Initial moles of HNO3 = 1Lx 0.083 mol HNO3/1 L
The HNO3 is a strong acid that can be completely dissociated in the following way.
HNO3→H + NO
1 mol 1 mol 1 mol
The HNO3 is dissociated into two particles (Kt and NO3). So that,
total moles particles = 0.083 mol (K) + 0.083 mol (NO) = 0.166 mol
The number of moles of solute = mass of solute ÷ molar mass of solute, where mass is measured in grams and molar mass (defined as the mass of one mole of a substance in grams) is measured in g/mol.
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