# Visha took two aqueous solutions — one containing 7·5 g of urea (Molar mass = 60 g/mol) and the other containing 42·75 g of substance Z in 100 g of water, respectively. It was observed that both the solutions froze at the same temperature. Calculate the molar mass of Z.

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

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

Hence, The molar mass (Mm) of Z = 342 g/mol

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