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2.22    At 300 \; K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98\; bar. If the osmotic pressure of the solution is 1.52\; bars at the same temperature, what would be its concentration?

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According to given conditions we have same solution under same temperature. So we can write :

                                                         \frac{\Pi _1}{C_1} = \frac{\Pi _2}{C_2}                                                   \left ( \Pi = C.R.T\ ; RT = \frac{\Pi }{C} \right )

So, if we put all the given values in above equation, we get

                                                \frac{4.98}{\frac{36}{180}} = \frac{1.52}{C_2}

or                                              C_2 = \frac{1.52\times36}{4.98\times180} = 0.061 M

Hence the required concentration is 0.061 M. 

 

                 

Posted by

Devendra Khairwa

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