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Elevation in Boiling Point

When 0.6 g ...

Question

When $0.6 g$ of urea dissolved in $100 g$ of water, the water will boil at $(K_{b} f o r w a t e r = 0.52 k J . m o l^{- 1}$ and
normal boiling point of water $= 100^{\circ} C)$ :

373.052 K

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273.52 K

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372.48 K

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273.052 K

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Solution

The correct option is B $373.052 K$
As we know,
$Δ T_{b} = \frac{1000 K_{b} w_{2}}{M_{2} w_{1}}$
$T_{b} - T_{b}^{\circ} = \frac{1000 \times 0.52 \times 0.6}{60 \times 100}$
$T_{b} - 373 = 0.052$
$T_{b} = 373.052$

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When 0.6g of urea dissolved in 100g of water, the water will boil at (Kbforwater=0.52kJ.mol−1and normal boiling point of water =100∘C):

The correct option is B 373.052KAs we know,ΔTb=1000Kbw2M2w1Tb−T∘b=1000×0.52×0.660×100Tb−373=0.052Tb=373.052

When $0.6 g$ of urea dissolved in $100 g$ of water, the water will boil at $(K_{b} f o r w a t e r = 0.52 k J . m o l^{- 1}$ and
normal boiling point of water $= 100^{\circ} C)$ :

The correct option is B $373.052 K$
As we know,
$Δ T_{b} = \frac{1000 K_{b} w_{2}}{M_{2} w_{1}}$
$T_{b} - T_{b}^{\circ} = \frac{1000 \times 0.52 \times 0.6}{60 \times 100}$
$T_{b} - 373 = 0.052$
$T_{b} = 373.052$