3. Two aqueous solutions containing respectively 7.5 g
urea (molar mass = 60) and 42.75 g substance X in
100 g of water freeze at the same temperature.
Calculate the molecular weight of X. •
Answers
Given:
The mass of water (w2) = 100 g
The mass of urea (w1) = 7.5 g
The mass of Z (w2) = 42.75 g
The molar mass of urea (Mm1) = 60 g/mol
Both the solutions have same freezing point.
To find:
The molar mass (Mm) of Z.
Solution:
Since both the solutions have the same freezing point the change in freezing points (ΔTf) for both the solutions is same.
ΔTf = Kf*m, hence both the solutions should have the same molality.
molality = moles of solute/ weight of solvent(in Kg)
Therefore, (w1*1000)/(Mm1*w2) = (w3*1000)/(Mm*w2) ⇒ Mm = w3*Mm1/w1 = 42.75*60/7.5 = 342 g/mol
Answer:
The molar mass (Mm) of Z = 342 g/mol
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Answer:
Given:
The mass of water (w2) = 100 g
The mass of urea (w1) = 7.5 g
The mass of Z (w2) = 42.75 g
The molar mass of urea (Mm1) = 60 g/mol
Both the solutions have same freezing point.
To find:
The molar mass (Mm) of Z.
Solution:
Since both the solutions have the same freezing point the change in freezing points (ΔTf) for both the solutions is same.
ΔTf = Kf*m, hence both the solutions should have the same molality.
molality = moles of solute/ weight of solvent(in Kg)
Therefore, (w1*1000)/(Mm1*w2) = (w3*1000)/(Mm*w2) ⇒ Mm = w3*Mm1/w1 = 42.75*60/7.5 = 342 g/mol
Answer:
The molar mass (Mm) of Z = 342 g/mol