A) \[C{{H}_{2}}O~\]
B) \[{{C}_{2}}{{H}_{4}}{{O}_{2}}\]
C) \[{{C}_{4}}{{H}_{8}}{{O}_{4}}\]
D) \[{{C}_{3}}{{H}_{6}}{{O}_{3}}\]
Correct Answer: B
Solution :
For isotonic solution (solution with same osmotic pressure) \[{{C}_{1}}={{C}_{2}}\] \[\frac{{{W}_{x}}}{{{M}_{x}}}=\frac{{{W}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}{{{M}_{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}}\] \[\frac{3}{{{M}_{x}}}=\frac{9}{180}\] \[0.05M\,{{C}_{6}}{{H}_{12}}{{O}_{6}}\]containing 9 g of glucose \[{{M}_{x}}=\frac{3\times 180}{9}=60\] Molecular weight of a non-electrolyte is 60 Molecular formula = n\[\times \] empirical formula Empirical formula \[=C{{H}_{2}}O\] Empirical formula mass \[=12+2+16=30\] \[n=\frac{molecular\,mass}{empirical\,formula\,mass}\] \[=\frac{60}{30}=2\] Molecular formula \[=2\times C{{H}_{2}}O={{C}_{2}}{{H}_{4}}{{O}_{2}}\]
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