When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass o
1 answer:
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>First, the equation of the reaction:
The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
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Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or )
Given : kf = 1.86
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907
)
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C