A) 19.0
B) 19.602
C) 19.402
D) 19.202
Correct Answer: B
Solution :
Moles of glucose; \[n{{C}_{6}}{{H}_{12}}{{O}_{6}}=0.1\] Moles of water; \[{{n}_{{{H}_{2}}O}}=\frac{178.2}{18}=9.9\] Mole fraction of water, \[{{x}_{{{H}_{2}}O}}=\frac{9.9}{9.9+0.1}=0.99\] \[{{P}_{solution}}={{P}_{solvent}}\times {{x}_{solvent}}\] \[=19.8\times 0.99\] \[=19.602mm\text{ }Hg\]You need to login to perform this action.
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