34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be

Question

34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be (A)  0.0099  (B)  1.1597  (C)  0.840  (D)  0.9901

Tardigrade Admin · Accepted Answer

(p°-pS/p°)=((w2/m2)/(w1)m1+(w2)M2 = ((34.2/342)/(34.2)342+(180)18 = (0.1/10.1) = 0.0099

Q. $34.2 g$ of cane sugar is dissolved in $180 g$ of water. The relative lowering of vapour pressure will be

$0.0099$

$1.1597$

$0.840$

$0.9901$

$\frac{p ^{\circ} - p _{S}}{p ^{\circ}} = \frac{\frac{w _{2}}{m _{2}}}{\frac{w _{1}}{m _{1}} + \frac{w _{2}}{M _{2}}}$

$= \frac{\frac{34.2}{342}}{\frac{34.2}{342} + \frac{180}{18}}$

$= \frac{0.1}{10.1}$

$= 0.0099$

Q. 34.2g of cane sugar is dissolved in 180g of water. The relative lowering of vapour pressure will be

0.0099

1.1597

0.840

0.9901