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

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Solution

The correct option is A $3$ g
$(Δ p)_{g l u c o s e} = (Δ p)_{u r e a}$
$(x_{B})_{g l u c o s e} = (x_{B})_{u r e a}$
i.e., ${(\frac{n_{B}}{n_{A}})}_{g l u c o s e} = {(\frac{n_{B}}{n_{A}})}_{u r e a}$
$\frac{w_{B}}{50} \times \frac{18}{180} = \frac{1 \times 18}{50 \times 60}$
$w_{B} = 3$ g.

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Similar questions

Q. The mass in grams of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as produced by dissolving 1 g of urea in the same quantity of water is:

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Q. 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?

Q. 18.2 g of urea is dissolved in 100g of water at

50^{o} C

. The lowering of vapour pressure produced is 5 mm Hg. Calculate the molecular mass of urea. The vapour pressure of water at

50^{o} C

is 92 mm Hg.

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

The correct option is A 3 g(Δp)glucose=(Δp)urea(xB)glucose=(xB)ureai.e., (nBnA)glucose=(nBnA)ureawB50×18180=1×1850×60wB=3g.

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

The correct option is A $3$ g
$(Δ p)_{g l u c o s e} = (Δ p)_{u r e a}$
$(x_{B})_{g l u c o s e} = (x_{B})_{u r e a}$
i.e., ${(\frac{n_{B}}{n_{A}})}_{g l u c o s e} = {(\frac{n_{B}}{n_{A}})}_{u r e a}$
$\frac{w_{B}}{50} \times \frac{18}{180} = \frac{1 \times 18}{50 \times 60}$
$w_{B} = 3$ g.