1292.5 gram of aqueous solution of 5 molal NaCl is kept in a large bucket. The bucket is placed under a tap from which 2 molal aqueous solution of NaCl is flowing. Rate of flow of solution from the tap is 0.5 gram/sec. The total amount of solution finally present in bucket, when concentration of NaCl solution in bucket is 4-molal, is found to be x gram and the time after which the bucket will have 4-molal concentration of NaCl is found to be y sec. Then-

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x= 1851gram

x= 2051 gram

y= 1117 sec

y= 1517 sec

answer is A, C.

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Detailed Solution

Let x-mole NaCl and Y kg of water is added from the tap in the bucket
Initially in the bucket
Mole of NaCl=5
Wt of water = 1kg
Finally, in the bucket
Mole of NaCl= (5+x)
Weight of water = $(1 + y) K . G$
Molality $= \frac{5 + x}{(1 + y)}$
$4 = \frac{5 + x}{(1 + y)}$
$4 + 4 y = 5 + x$
$4 y - x = 1 ....... (1)$
Also $2 = \frac{x}{y}$
   $x = 2 y$
   $2 y = 1$
$y = \frac{1}{2}$
   $x = 1$
The total amount of solution finally present $= 1292.5 + 1 \times 58.5 + \frac{1}{2} \times 1000$
$= 1851 g r a m$
Time required $= \frac{558.5}{0.5}$
$= 1117 \sec$