A) \[300\]
B) \[600\]
C) \[900\]
D) \[1200\]
Correct Answer: C
Solution :
Molarity of glucose solution \[=\frac{0.5g}{180\times 1lit.}\] \[=0.0027M\] Isotonic solution have equal molar concentration, hence molarity of organic compound solution \[=0.0027\text{ }M\] \[\therefore \] Molecular weight of compound \[=\frac{2.5}{0.0027\times 1}\] \[=900\]You need to login to perform this action.
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