200 mL of 0.5 M, 300 mL of 0.7 M and 500 ml of 2 M solutions of ethanol are mixed. The final solution is diluted by 10 times. Its molarity is:

1.31 M

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13.1 M

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0.131 M

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131 M

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Solution

The correct option is C 0.131 M
$N o . o f m o l e s = m o l a r i t y \times v o l u m e (i n l i t r e$
Hence,
$N o . o f m o l e s i n 200 m L 0.5 M = 0.5 \times \frac{200}{1000}$ $= 0.1$
$N o . o f m o l e s i n 300 m L 0.7 M = 0.7 \times \frac{300}{1000}$ $= 0.21$ $N o . o f m o l e s i n 500 m L 2.0 M = 2.0 \times \frac{500}{1000}$ $= 1.0$
Now,
$T o t a l n o . o f m o l e s = 0.1 + 0.21 + 1 = 1.31$
$T o t a l V o l u m e = (200 + 300 + 500) m L = 1000 m L = 1 L$
$M o l a r i t y o f s o l u t i o n a f t e r m i x i n g = \frac{1.31}{1} = 1.31 M$
After 10 times dilution,
Molarity = 0.131 M

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