The VP of a solution of 5g of non - electrolyte in 100 g of $H_{2} O$ at a particular temp is 2985 $N m^{- 2}$ . The VP of pure water at that temperature is 3000 $N m^{- 2}$ . The molecular weight of the solute is:

180

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

270

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 180
$M_{2} = \frac{W_{2} M_{1}}{W_{1}} (\frac{p^{\circ}}{p^{\circ} - p})$
$= \frac{5 \times 18}{100} [\frac{3000}{15}] = 180$

Suggest Corrections

Similar questions

Vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is 2985 N/m². The vapour pressure of pure water is 3000 N/m². The molecular weight of the solute is:

Q. The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is

2985 N m^{- 2}

. The vapour pressure of pure water at that temperature is

3000 N m^{- 2}

Q. The vapour pressure of a solution containing 5 g of a non electrolyte in 100 g of water at a particular temperature is

2985 N m^{- 2}

. If the vapour pressure of pure water is

3000 N m^{- 2}

, the molecular weight of solute is:

Q. The vapour pressure (VP) of a dilute solution of non-volatile solute is P and the VP of a pure solvent is

P^{o}

. The lowering of the VP is:

Q. The vapour pressure of pure benzene at a particular temperature is

260

bar. At that temperature, the vapour pressure of a solution containing

2

g non-volatile non-electrolytic solid in

100

g of benzene is

250

bar. What is the molecular mass of the solute?

Join BYJU'S Learning Program

The VP of a solution of 5g of non - electrolyte in 100 g of H2O at a particular temp is 2985 Nm−2. The VP of pure water at that temperature is 3000 Nm−2. The molecular weight of the solute is:

The correct option is A 180M2=W2M1W1(p∘p∘−p) =5×18100[300015]=180

The VP of a solution of 5g of non - electrolyte in 100 g of $H_{2} O$ at a particular temp is 2985 $N m^{- 2}$ . The VP of pure water at that temperature is 3000 $N m^{- 2}$ . The molecular weight of the solute is:

The correct option is A 180
$M_{2} = \frac{W_{2} M_{1}}{W_{1}} (\frac{p^{\circ}}{p^{\circ} - p})$
$= \frac{5 \times 18}{100} [\frac{3000}{15}] = 180$