The mass of glucose that should be dissolved in $100 g$ water in order to produce same lowering of vapour pressure as produced by dissolving $1 g$ urea in $50 g$ of water is?

1 g

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2 g

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6 g

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12 g

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Solution

The correct option is C $6 g$
$(H_{2} N C O N H_{2}) \to M = 60 g / m o l$
$m =$ molar mass
$M_{C_{6} H_{12} O_{6}} = 12 \times 6 + 12 \times 1 + 6 \times 6 = 72 + 12 + 96 + 180 g / m o l$
Relative lowering in Vapour pressure $= n o . o f m o l e s \times \frac{m o l a r m a s s o f s o l v e n t}{m a s s o f w a t e r} = 0.33 \times \frac{18}{50}$
$\frac{W}{180} \times \frac{18}{100} = \frac{1}{60} \times \frac{18}{50} W = 6 g$

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The mass of glucose that should be dissolved in 100g water in order to produce same lowering of vapour pressure as produced by dissolving 1g urea in 50g of water is?

The correct option is C 6g(H2NCONH2)→M=60g/molm= molar massMC6H12O6=12×6+12×1+6×6=72+12+96+180g/molRelative lowering in Vapour pressure=no.ofmoles×molarmassofsolventmassofwater=0.33×1850W180×18100=160×1850W=6g

The mass of glucose that should be dissolved in $100 g$ water in order to produce same lowering of vapour pressure as produced by dissolving $1 g$ urea in $50 g$ of water is?