500g if water containing 27 g of non volatile solute will boul at 100.156° C. calculate the molar mass of the solute. give boiling point of water= 100°C , Kb= 0.52K kg mol^ -1.
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180 g/mole
Explanation:
Given: Elevation in boiling point = ΔTb = 100.156 - 100 = 0.156 °C
mass of solute = x g/mole , Kb= 0.52K kg mol^ -1.
Number of moles of solute = 27/x
Molality = m= number of moles of solute × 1000/mass of water
= 27 × 1000/x × 500
= 54/x
We know that,
ΔTb = Kb.m
0.156 = 0.52 × 54/x
x = 0.52 × 54/0.156
= 180 g/mole
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