The mass of a non-volatile solute of molar mass 60 g mol -1 that should be dissolved in 126 g of water to reduce its vapour pressure nto 99 % wil be

Question

The mass of a non-volatile solute of molar mass 60 g mol -1 that should be dissolved in 126 g of water to reduce its vapour pressure nto 99 % wil be (A) 4 g (B) 8 g (C) 5.6 g (D) 3 g

Tardigrade Admin · Accepted Answer

Relative lowering in vapour pressure = mole fraction of the non-volatile solute (p°-ps/p°)=x2=(w2 / M2/w1 / M1+w2 / M2) (100-99/100)=(w2 / 60/126 / 18+w2 / 60) w2=4.2=4 g

Q. The mass of a non-volatile solute of molar mass $60 g m o l^{- 1}$ that should be dissolved in $126 g$ of water to reduce its vapour pressure nto $99%$ wil be

4 g

8 g

5.6 g

3 g

Relative lowering in vapour pressure
= mole fraction of the non-volatile solute
$\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = x_{2} = \frac{w _{2} / M _{2}}{w _{1} / M _{1} + w _{2} / M _{2}}$
$\frac{100 - 99}{100} = \frac{w _{2} /60}{126/18 + w _{2} /60}$
$w_{2} = 4.2 = 4 g$

Q. The mass of a non-volatile solute of molar mass 60gmol−1 that should be dissolved in 126g of water to reduce its vapour pressure nto 99% wil be

4 g

8 g

5.6 g

3 g

Relative lowering in vapour pressure = mole fraction of the non-volatile solute p∘p∘−ps​​=x2​=w1​/M1​+w2​/M2​w2​/M2​​ 100100−99​=126/18+w2​/60w2​/60​ w2​=4.2=4g

Q. The mass of a non-volatile solute of molar mass $60 g m o l^{- 1}$ that should be dissolved in $126 g$ of water to reduce its vapour pressure nto $99%$ wil be

Relative lowering in vapour pressure
= mole fraction of the non-volatile solute
$\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = x_{2} = \frac{w _{2} / M _{2}}{w _{1} / M _{1} + w _{2} / M _{2}}$
$\frac{100 - 99}{100} = \frac{w _{2} /60}{126/18 + w _{2} /60}$
$w_{2} = 4.2 = 4 g$