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2.40    Determine the amount of  CaCl_{2} (i=2.47)  dissolved in 2.5\; litre of water such that its osmotic pressure is 0.75 \; atm at  27^{\circ}C.

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We know that osmotic pressure :

                                                    \Pi = i\ (\frac{n}{v})\ R\ T

or                                                 \Pi = i\ (\frac{w}{M\ v})\ R\ T

We have been given the values of osmotic pressure, V, i and T.

So the value of w can be found.

                                                  w = \frac{0.75\times111\times2.5}{2.47\times0.0821\times300}                                                         (M = 1\times40 + 2\times 35.5 = 111\ g\ mol^{-1})

                                                        = 3.42\ g

Hence 3.42 g CaClis required.

Posted by

Devendra Khairwa

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