EXPERT PLS SEND ANSWER URGENTLY!!! Share with your friends Share 0 Aditi Gupta answered this Solution: Moles of sugar = 1 mol Mass of sugar = Mole×Molar mass = 1 × 342 = 342 gMoles of water = 3 molMass of water = 3 × 18 = 54 gRelative lowering of vapour pressure = P10-PP10 = W2/M2W1/M1+W2/M2 = 342/342342/342 + 54/18 = 11 + 3 = 0.25 Hence, the correct option is (1). 0 View Full Answer
