A) 16.500 mm Hg
B) 17.325 mm Hg
C) 17.675 mm Hg
D) 15.750 mm Hg
Correct Answer: B
Solution :
Moles of glucose \[=\frac{18}{180}=0.1\], moles of \[{{H}_{2}}O=\frac{1782}{18}=9.9\] \[\frac{{{P}^{o}}-{{P}_{S}}}{{{P}_{S}}}=\frac{moles\text{ }of\text{ }glucose}{moles\text{ }of\text{ }water}\,\, & ;\,\,\,17.5-{{P}_{S}}\]\[=\frac{0.1\times {{P}_{S}}}{9.9}\] \[{{P}_{S}}=17.325\] mm Hg.You need to login to perform this action.
You will be redirected in
3 sec