The boiling point of a solution containing 50g of a non volatile solute in 1kg of solvent is 0.5^oC higher than the pure solvent. Determine the mass of the solute.
K_b for solvent = 2.53 K KG mol^-1
Molecular mass of solvent = 78 g mol^-1

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Solution

Dear student,

ΔT_b = K_b m

n (moles of solute) = $\frac{50}{M o l a r m a s s}$

molality (m) = $\frac{n}{1} = \frac{n m o l e s}{k g}$

0.5 = 2.53 $\frac{n}{1}$

n = 1.976

1.976 = $\frac{50}{m o l a r m a s s}$

molar mass of solute= $\frac{50}{1.976}$ = 25.3 g/mol (Ans)

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