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ChemistryLiquid SolutionMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

The mass (in grams) of glucose that should be dissolved in 100 g of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 g of urea (molecular mass = 60) in 50 g of water is

Answer:

6.00

Solution:

$\frac{Δ P}{P_{0}} = X_{B}$
$\approx \frac{n_{B}}{n_{A}} = \frac{w_{B} \times m_{A}}{m_{B} \times w_{A}}$
${(\frac{w_{B} \times m_{A}}{m_{B} \times w_{A}})}_{g l u c o s e} = {(\frac{w_{B} \times m_{A}}{m_{B} \times w_{A}})}_{u r e a}$
$\frac{w_{B} \times 18}{180 \times 100} = \frac{1 \times 18}{60 \times 50}$
$w_{B} = 6$ g.

Stream:NTA_ABHYASSubject:ChemistryTopic:Liquid SolutionSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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