The vapour pressure of 100 g water reduces from 3000 Nm^ -2 to 2985 Nm^ -2 when 5 g of substance ' X ' is dissolved in it. Substance ' X ' is

Correct answer: B. glucose aligned p^ -p_s p^ =(n)/(n+N) & =((w)/(m))/((w)/(m)+(W)/(M)) (w)/(m) & (W)/(M) p^ -p_s p^ & =(w / m)/(W / M)=(w)/(m) × (M)/(W) (3000-2985)/(3000) & =(5)/(m) × (18)/(100) m & =(5 × 18 × 3000)/(15 × 100) m & =180 aligned 180 is the molecular weight of glucose (C_6 H_ 12 O_6) , thus the substance ' X ' is glucose.

JIPMER 2010

Exam: JIPMER
Year: 2010
Subject: Chemistry
Chapter: Solutions

JIPMER 2010 · Chemistry · Solutions

The vapour pressure of 100 g water reduces from 3000 Nm^ -2 to 2985...

The vapour pressure of 100 g water reduces from

3000 \mathrm{Nm}^{-2}

2985 \mathrm{Nm}^{-2}

when 5 g of substance '

X

' is dissolved in it. Substance '

X

' is

Options

Amethanol
Bglucose
Ccarbon dioxide
DCannot predict

Correct answer

B. glucose

Step-by-step solution

\begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n}{n+N} & =\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}} \\ \because \quad \frac{w}{m} & \lt \frac{W}{M} \\ \therefore \quad \frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w / m}{W / M}=\frac{w}{m} \times \frac{M}{W} \\ \frac{3000-2985}{3000} & =\frac{5}{m} \times \frac{18}{100} \\ m & =\frac{5 \times 18 \times 3000}{15 \times 100} \\ m & =180\end{aligned}

180 is the molecular weight of glucose

\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)

, thus the substance '

X

' is glucose.

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