A) 300
B) 600
C) 925
D) 1200
Correct Answer: C
Solution :
Given, Glucose solution \[\text{=}\,\text{0}\text{.5}\,\text{g/L}\]is isotonic with \[\text{2}\text{.5 g/L}\]solution of organic compound. Molarity of glucose \[=\frac{\text{mass}\,\text{of}\,\text{glucose/molecular}\,\text{mass}\,\text{of}\,\text{glucose}}{\text{V}\,\text{of}\,\text{solution}\,\text{in}\,\text{L}}\]\[=\frac{0.5/180}{1}=0.0027\,M\] \[\because \]Glucose solution is isotonic with organic compound solution. \[\therefore \]molarity of glucose = molarity of organic compound \[\text{= 0,0027 M}\] Molarity of organic compound mass of organic compound/ \[=\frac{\text{molecular}\,\text{mass}\,\text{of}\,\text{organic}\,\text{compound}}{\text{V}\,\text{of}\,\text{solution}}\] or \[0.0027=\frac{\text{2}\text{.5/molecular}\,\text{mass}}{\text{1}}\] \[\therefore \] \[\text{molecular}\,\text{mass}=\frac{2.5}{0.0027}\] \[=925\]
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